Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Students can Download Chapter 4 Chemical Bonding and Molecular Structure Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Plus One Chemistry Chemical Bonding and Molecular Structure One Mark Questions and Answers

Question 1.
The octet rule is not valid for
a) CO2
b)H2O
c) O2
d) CO
Answer:
d) CO

Question 2.
The stability of an ionic crystal depends principally on
a) High electron gain enthalpy of the anion forming species
b) The lattice enthalpy of the crystal
c) Low ionization enthalpy of the cation forming species
d) Low heat of sublimation of cation forming solid
Answer:
b) The lattice enthalpy of the crystal

Question 3.
Which of the following molecules has highest dipole moment?
a) H2S
b)CO2
c) CCl4
d) BF3
Answer:
a) H2S

Question 4.
The d-orbital involved in sp3d hybridization is .
Answer:
dz2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 5.
Which of the following is paramagnetic and has a bond order of ½?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 1
Answer:
H2+

Question 6.
Dipole moment µ electric charge ‘e’ and bond length ‘d’are related by the equation.
Answer:
M = e x d

Question 7.
In which of the following carbon atom is sp2 hybridised?
a) CO2
b) C2H6
c) C6H6
d) HCN
e) \(C{ H }_{ 3 }-C\equiv CH\)
Answer:
C6H6

Question 8.
AgF is ionic whereas Agcl is covalent. This can be explained by
Answer:
Faja’ens Rule

Question 9.
The shape of covalent molecule CIF3 is _________
Answer:
T – shaped

Question 10.
The C – O bond order in CO32- is
Answer:
1.33

Plus One Chemistry Chemical Bonding and Molecular Structure Two Mark Questions and Answers

Question 1.
The order of repulsion of electron pairs as written by student is given below:
lone pair-lone pair repulsion < lone pair-bond pair repulsion>bond pair-bond pair repulsion.
1. Can you see anything wrong in this?
If yes, correct it.
2. Name the theory behind this.
Answer:
1. Yes.
Repulsion decreases in the order: lone pair-lone pair repulsion>lone pair-bond pair repulsion> bond pair-bond pair repulsion,

2. VSEPR theory

Question 2.
During a small group discussion in the class room a student argued that in acetylene both the carbon atoms are in sp3 hybridised state.

  1. What is your opinion?
  2. What is the bond angle between carbon atoms in acetylene?

Answer:

  1. The student’s argument is wrong. In acetylene both the carbon atoms are triple bonded and are in sp hybridised state,
  2. 180°

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Classify the following compounds according to their hybridisation.
CH4, BF3, C2H4, BeF2, C2H2
Answer:
sp3 hybridisation – CH4
sp2 hybridisation – BF3, C2H4
sp hybridisation – BeF2, C2H2

Question 4.
A student arranged the halide ions in the increasing order of polarisability as: F < l < CI < Br
1. Is this the correct order? If not write it in correct order.
2. Justify.
Answer:
1. No.
Polarisability increases in the order: F < Cl <Br < l

2. Polarisability increases when the size of anion increases.

Question 5.
Give any two differences between sigma and pi bonds.
Answer:
Sigma bond (σ) is formed when two atomic orbitals under head-on overlapping. It is a strong bond. Pi(π) bond is formed when atomic orbitals undergo lateral (sidewise) overlapping. It is a weak bond.

Question 6.
Write the type of hybridisation of each carbon in the compound.
CH3-CH=CH-CN
Answer:
Carbon 1 → sp³
Carbon 2 → sp²
Carbon 3 → sp²
Carbon 4 → sp

Plus One Chemistry Chemical Bonding and Molecular Structure Three Mark Questions and Answers

Question 1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 2
1. What is meant by the above picture?
2. Which type of bond is present here?
3. Which type of overlapping leads to the formation of π bond?

Answer:

  1. s-s overlapping
  2. A strong sigma bond
  3. This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals.

Question 2.
‘In ethane there are 6 covalent bonds. Five are strong σ bonds and the remaining one is a weak π bond.’

  1. Do you agree with this?
  2. How is a bond different from π bond in the mode of formation?

Answer:

  1. Yes.
  2. Sigma bond is formed by the end to end overlapping of bonding orbitals along the internuclear axis, π bond is formed by the lateral or sidewise overlap of half filled atomic orbitals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Choose the correct molecules from the given clues: H2O, SF6, BF3
1. Clue -1 The central atom is in sp² hybridised state and the molecule has trigonal planar in shape. Clue-2 The bond angle is 120°
2. Clue-1 The number of electron pairs in this molecule is 6.
Clue -2 It has octahedral geometry.
3. Clue-1 The bond angle is reduced from 109° 28′ to 104.5°
Clue-2 It has a bent shape.
Answer:

  1. BF3
  2. SF6
  3. H2O

Question 4.
Give theoretical explanation for the following statements.
1. H2S is acidic while H2O is neutral.
2. Hydrogen chloride gas dissolves in water.
Answer:
1. S-H bond energy is less than that of O-H bond energy. So H+ can be easily generated from H2S.

2. When HCl is treated with H2O it undergoes hydrolysis as per the following reaction and dissolves.
HCl + H2O → H3O+ +Cl

Question 5.
The potential energy level diagram forthe formation of hydrogen molecule as drawn by a student is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 3
1. Resketch the graph with correct labelling.
2. How can you determine the radius of one hydrogen atom?
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 4
2. Bond length in hydrogen molecule is the intermolecular distance between two hydrogen atoms. So, half of the bond length is taken as the radius of hydrogen atom.

Question 6.
Ionisation enthalpy is one of the factors favoring the
formation of ionic bonds.
1. Will you agree with this statement?
2. Explain how?
3. Write anotherfactorfavouring the formation of ionic bonds.
Answer:
1. Yes.

2. In the formation of the ionic bond, a metal atom losses electrons to form cation. This process requires energy equal to the ionisation enthalpy. Lesser the ionisation enthalpy of the metal atom, easier will be the removal of electron from the atom to form cation and hence greater will be the tendency to form ionic bond.

3. Electron gain enthalpy of the element forming anion.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 7.
Complete the following table:

Sigma bond π -bond
Formation …………. ………….
Strong/Weak …………. ………….
About rotation …………. ………….

Answer:

Sigma bond π -bond
Formation Sigma bond is formed by end to end (or axial) overlap of atomic orbitals π -bond is formed by the sidewise overlap of atomic orbitals
Strong/Weak Strong Weak
About rotation Free rotation Free rotation is not possible

Question 8.
a) The dipole moment of BF3 is zero even though the B – F bonds are polar. Justify.
b) Give the hybridisation involved in the following compounds

  1. NH3
  2. C2H4
  3. SF6
  4. PCl5

c) o-nitro phenol has a lower boiling point than its para isomer. Why?
Answer:
a) In BF3, dut to the symmetric trigonal planar geometry of the molecule, the B – F bond are oriented at an angle of 120°to one another. The three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 5
b)

  1. sp³
  2. sp²
  3. sp³d²
  4. sp³d

c) In o-nitrophenol, intramolecular hydrogen bonding is present and there is no association of molecules whereas in p-nitrophenol there is inter-molecular hydrogen bonding which causes association of molecules.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 6

Question 9.
1. How many a and n bonds are there in the following molecules i) ethane ii) acetylene?
2. BF3 and NH3 are tetra atomic molecules. But the shape of BF3 is different from that of NH3. Explain this using hybridisation.
Answer:
1. i) Ethane – 7σ bonds
ii) Acetylene-3σ bonds and 2 π bonds

2. In BF3 molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital. As a result, boron has three unpaired electrons. These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals oriented in a trigonal planar arrangement and overlap with 2 p orbitals of F to form three B – F bonds. Therefore, BF3 molecule has a planar geometry with FBF bond angle of 120°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 7
In NH3, the valence shell electronic configuration of N in ground state is 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\). These four orbitals undergo sp³ hybridisation to form four sp³ hybrid orbitals, three of them containing unpaired electrons and the fourth one containing lone pair. The three hybrid orbitals overlap with 1 s orbitals of hydrogen atoms to form three N – H sigma bonds. Since, the bp-lp repulsion is greater than the bp-bp repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. Thus, the geometry of NH3 molecule is trigonal pyramidal.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 8

Question 10.
Covalent bond is formed by the overlaping of atomic orbitals.
1. What is meant by orbital overlapping?
2. What are the 3 types of overlapping?
Answer:
1. Orbital overlapping is the partial interpenetration or merging of atomic orbitals. It results in the pairing of electrons. Greater the overlap the stronger is the bond formed between two atoms.

2. s-s overlapping
s-p overlapping
p-p overlapping

Question 11.
1. Which among the following will exist He2 or He2+? Explain.
2. H2S is a gas at ordinary condition, while H2O is liquid. Account for the above statement.
3. State the hybridisation in the following molecules,
i) PF6
ii) C2H6
Answer:
1. He2+
Helium molecule contains 4 electrons. Out of this 4 electrons, 2 are present in the bonding molecular orbital and the remaining 2 are present in the anti-bonding molecular orbital.
Bond order = ½ (Nb-Na)
= ½ (2-2) = 0
Hence, He2 cannot exist. The molecular orbital diagram is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 9
He2+ contains 3 electrons. Out of these 3 electrons, 2 are present in the σ1s level and the remaining one is present in the σ* 1s level.
Bond order = ½ (Nb-Na)
= ½ (2-1) = ½
Since the bond order is half the molecular ion exists but possesses low stability.

2. In H2S, there is no hydrogen bonding whereas in water, molecular association is possible due to intermolecular hydrogen bonding.

3. i) PF5 = sp³d hybridisation
ii) C2H6 = sp³ hybridisation

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 12.
Bond order is a term commonly used in MO theory.
1. How is it calculated?
2. How is it related to bond length and bond energy?
Answer:
1. Bond order = ½ (Nb-Na)
where Nb = No. of electrons occupying bonding orbitals and Na = No. of electrons occupying antibonding orbitals.

2. As the bond order increases, bond length decreases and bond energy increases, i.e., bond order is directly proportional to bond energy and inversely proportional to bond length.

Question 13.
1. Explain the hybridisation and geometry of ethyne.
2. What is the difference between bonding molecular orbital and antibonding molecular orbital?
3. How the magnetic nature of a molecule is related to its electronic structure?
Answer:
1. In the formation of ethyne (C2H2), both the carbon atoms undergo sp hybridisation having two unhybridised orbitals (2px and 2py). One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming o bonds. Each of the two unbybridised p orbitals of both the carbon atoms overlaps sidewise to form two K bonds between the carbon atoms. Thus, ethyne has a linear geometry with π bond angle of 180°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 10

2. The molecular orbital which has lower energy than the atomic orbital is called bonding molecular orbital and the molecular orbital which has greater energy than the atomic orbital is called anti bonding molecular orbital.

3. If all the molecular orbitals in a molecule are doubly occupied (i.e., paired), the substance is diamagnetic (repelled by magnetic field). It one or more molecular orbitals are singly occupied (i.e., unpaired) it is paramagnetic (attracted by magnetic field).

Question 14.
Molecular Orbital Theory (MOT) is an advanced theory
of chemical bonding.
1. Write the salient features of MOT.
2. What is meant by LCAO? Illustrate using hydrogen molecule.
3. What are the conditions for the combination of atomic orbitals?
Answer:
1.

  • The electrons in a molecule are present in the
    various molecular orbitals.
  • The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
  • iii) The electron in a molecular orbital is influenced by two or more nuclei depending upon the number of atoms in the molecule.
  • The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two
    molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
  • The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
  • The electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
  • The molecular orbitals are filled in accordance with the Aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule.

2. LCAO refers to the linear combination of atomic orbitals. It is an approximate method used to explain the formation of molecular obritals. Consider hydrogen molecule consisting of two atoms A and B. Each hydrogen atom has one electron in the 1s orbital. The atomic orbitals of these atoms can be represented by the wave functions ψA and ψB. Mathematically, the formation of molecular orbitals can take place by addition and by subtraction of wave functions of individual atomic orbitals.
ψMO = ψA ± ψB
Therefore, the two molecular orbitals σ and σ* are formed as:
σ* = ψA – ψB
The molecular orbital σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ* formed by the subtraction of atomic orbital is called antibonding molecular orbital. The energy level diagram is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 11

3.

  • The combining atomic orbitals must have the same or nearly same energy.
  • The combining atomic orbitals must have the same symmetry about the molecular axis.
  • The combining atomic orbitals must overlap to the maximum extent.

Question 15.
Consider a reaction PCl5(g) → PC3(g) + Cl2(g)
1. What is the change in hybridisation state of phosphorus?
2. Explain why does PCl5 decomposes easily?
Answer:
1. When PCl5 decomposes to PCl3, the hybridisation of P changes from sp³d to sp³.

2. In PCl5, the five sp³d orbitals of P overlap with the singly occupied p orbitals of Cl atoms to form five P-CI sigma bonds. Three P-Cl bonds which lie in one plane and make an angle of 120° with each other are called equatorial bonds. The remaining two P – Cl bonds, called axial bonds, one lie above and the other lie below the equatorial plane, make an angle of 90° with the plane. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, axial bonds are slightly longer and hence slightly weaker than the equatorial bonds. This makes PCl5 molecule more reactive and hence it decomposes easily.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 12

Question 16.
The electron dot structure (Lewis structure) of ammonia molecule is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 13
1. Write the number of bond pairs of electrons and lone pairs of electrons in ammonia molecule.
2. The structures of o-nitrophenol and p-nitrophenol are shown in the figure. The former is a steam volatile liquid whereas the latter is a solid. Justify your answer giving reason.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 14
Answer:
1. Ammonia molecule contains one lone pair of electrons and 3 bond pair of electrons,

2. In o-nitrophenol, there is intramolecular hydrogen bonding and there is no molecular association. But in p-nitro phenol intermolecular hydrogen bonding is present and hence molecular association is possible.

Plus One Chemistry Chemical Bonding and Molecular Structure Four Mark Questions and Answers

Question 1.
Hydrogen bonding is present in NH3 and H2O.
1. What is hydrogen bond?
2. What are different types of hydrogen bonds?
3. Explain the effect of hydrogen bonding.
Answer:
1. Hydrogen bond is defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O, N) of the same or another molecule.

2. Intermolecular hydrogen bond and Intramolecular hydrogen bond.

3. Compounds containing hydrogen bonds show higher melting and boiling points. Compounds whose molecules can form hydrogen bonds with water molecules are soluble in water.

Question 2.
Classify the following compounds according to their
shape.
BeF2, BeCl2, CH4, BF3, PCl5, SF6, SbCl5 NH4+, SiF4, AlCl3.
Answer:
Linear – BeF2, BeCl2
Trigonal planar-AlCl3, BF3
Tetrahedral – CH4, NH4+, SiF4
T rigonal bipyramidal – PCl5, SbCl5

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Benzene is an example of a compound exhibiting resonance.
1. What is meant by resonance?
2. Explain the resonance of ozone.
Answer:
1. When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more different structures, then the actual molecule is said to be a resonance hybrid of these canonical structures.

2. The resonance in ozone can be represented by the following structures:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 15
According to structures I and II, there is one single bond and one double bond in ozone molecule. But experiments show that both the oxygen- oxygen bonds are equal and the bond length 128 pm) is intermediate between single (148 pm) and double bond (121 pm) lengths. Hence it is assumed that ozone is a resonance hybrid (structure III) of structures I and II.

Question 4.
Match the following:

No. of electrons pairs Shape of molecule Examples
2 Trigonal planar SF6
4 Linear BeF2, BeCl2
3 Tetrahedral BF3, AlCl3
6 Octahedral CH4, SiF4

Answer:

No. of electrons pairs Shape of molecule Examples
2 Linear BeF2, BeCl2
4 Tetrahedral CH4, SiF4
3 Trigonal planar BF3, AlCl3
6 Octahedral SF6

Question 5.
In the formation of methane, carbon undergoes sp³ hybridisation.

  1. What do you mean by sp³ hybridisation?
  2. Give the % s-character and p-character of an sp³ hybrid orbital.
  3. What is the bond angle in methane?
  4. What is the geometry of methane molecule?

Answer:

  1. sp³ hybridisation involves mixing up of one – s and three-p orbitals of the valence shell of an atom to form four sp³ hybrid orbitals of equivalent energies and shape.
  2. Each sp³ hybrid orbital has 25% s-character and 75% p-character.
  3. The angle between the sp3 hybrid orbitals in methane is 109°28′.
  4. Tetrahedron.

Question 6.
Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 16
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 17

Question 7.
Fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 18
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 19

Question 8.
Dipole moment is used to predict the shape of
molecules.
1. Justify the statement based on the shapes of CO2 and H2O.
2. Which is having high dipole moment? NH3 or NF3? Why?
Answer:
1. Carbon dioxide is a linear molecule in which the two C=0 bonds are oriented in the opposite directions at an angle of 180°. Hence the two C=0 bond dipoles cancel each other and the resultant dipole moment of CO2 is zero. Thus CO2 is non¬polar molecule
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 20
On the other hand, water molecule has a bent structure in which two O-H bonds are oriented at an angle of 104.5°. Therefore, the bond dipoles of two O-H bonds do not cancel each other and the molecule will have a net dipole moment (1.85D).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 21

2. The dipole moment of NH3 is higher than that of NF3. In both cases, the central N atom has a lone pair whose orbital dipole points away from the N atom. In NH3 the orbital dipole due to the lone pair is in the same direction as the resultant bond dipole of the three N-H bonds. On the other hand, in the case of NF3, the resultant dipole of the three N-F bonds is in the opposite direction to the orbital dipole due to the lone pair. Thus, the orbital dipole due to the lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 22

Question 9.
The geometry of a covalent molecule is related to the hybridisation involved in the central atom. Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 23
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 24

Question 10
Depending upon the type of overlapping, covalent bonds are of two types.
a) Name them and give any two difference between them.
b) Find the total number of these two types of bonds ’ in propane and 2-butene.
Answer:
a) Sigma (σ) bond and pi (π) bond.
Sigma (σ) bond:
This type of covalent bond is formed by the end to end overlapping of half-filled atomic orbitals along the internuclear axis. The overlap is also known as head on overlap or axial overlap. The electrons constituting sigma bond are called sigma electrons.

Pi (π) bond:
This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals. The atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicularto the internuclear axis.
b) 10 σ bond in propane
11 σ bond and 1π bond in 2-butene

Plus One Chemistry Chemical Bonding and Molecular Structure NCERT Questions and Answers

Question 1.
Explain the formation of a chemical bond. (2)
Answer:
According to Kossel-Lewis approach, the formation of chemical bond between the two atoms takes place either by the transference of electrons or by mutual sharing of electrons. However, according to the modem view the formation of chemical bond between the two approaching atoms occurs only if there is a net decrease of energy because of attractive and repulsive forces.

Question 2.
Write the favourable conditions for the formation of ionic bond. (2)
Answer:
Ionic bond is formed by transference of electrons from one atom to another. The favourable conditions for its formation are:

  • Low ionisation enthalpy of element forming cation.
  • More negative value of electron gain enthalpy of element forming the anion and
  • High value of lattice enthalpy of the compound formed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that in ammonia. Discuss. (2)
Answer:
The difference in bond angles is due to the different numbers of lone pairs and bond pairs in the two species. In NH3, the N atom has two lone pairs and three bond pairs while in H2O, the O atom has two lone pairs and two bond pairs. The repulsive interactions of lone pairs and bond pairs in water are relatively more than those in NH3. Hence, bond angle around central atom in water is relatively smaller (104.5°) than that in NH3 molecule (107°).

Question 4.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? (2)
BF3 + NH3 → F3B.NH3
Answer:
During combination of species BH3 and NH3, N atom of NH3 is donor and B atom of BF3 is acceptor. The hybrid state of B in BF3 is sp² and that of N in NH3 is sp³. In the compound F3B+-NH3 both N and B atoms are surrounded by four bond pairs. Thus, the hyrid state of both is sp³. Hence during the reaction the hybrid state of B changes from sp² to sp³ but that of N remains the same.

Question 5.
Define hydrogen bond. Is it weaker or stronger than the van der Waals’ forces? (2)
Answer:
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Hydrogen bond is stronger than van der Waals’forces because it is a strong dipole-dipole interaction. The van der Waals’ forces, on the other hand, are weak dispersion forces.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Students can Download Chapter 6 Thermodynamics Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Plus One Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
Hot coffee in a thermos flask is an example of system.
Answer:
Isolated

Question 2.
Which of the following statements is incorrect about internal energy?
a) The absolute value of internal energy cannot be determined
b) The internal energy of one mole of a substance is same at any temperature or pressure
c) The measurement of heat change during a reaction by bomb calorimeter is equal to the internal energy change
d) Internal energy is an extensive property
Answer:
b) The internal energy of one mole of a substance is same at any temperature or pressure

Question 3.
For which of the following the standard enthalpy is not zero?
a) C (Diamond)
b) C (Graphite)
c) Liquid mercury
d) Rhombicsulphur
Answer:
a) C (Diamond)

Question 4.
Say TRUE or FALSE?
Any spontaneous process must lead to a net increase in entropy of the universe.
Answer:
TRUE

Question 5.
The ∆H fora reaction is-30 kJ. On the basis of this fact, we can conclude that the reaction
a) Gives off thermal energy
b) Is fast
c) Is slow
d) Is spontaneous
Answer:
a) Gives off thermal energy

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 6.
Write the type of system in each of the following:

  1. Hot water taken in an open vessel
  2. Hot water taken in a closed metallic vessel
  3. Hot water taken in a thermos flask

Answer:

  1. Open system
  2. Closed system
  3. Isolated system

Question 7.
In a reversible process the total change in entropy is ∆s(universe) is
Answer:
Zero

Question 8.
For the reaction Ag2O \(\rightleftharpoons \) 2Ag + \(\frac{1}{2}\)O2(g) ∆S and ∆H are 66J K-1mol-1 and 30.56 Kg mol respectively. The reaction will not be spontaneous at.
Answer:
463K

Question 9.
One mole of methane undergoes combustion to form CO2 and water at 25°C. The difference between ∆U & ∆H will be
Answer:
-2RT

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
A gas expands from 1 l to 6 l against a constant pressure of 1 atm and it absorbs 500J of heat ∆μ is
Answer:
-6.5J

Question 11.
Born Haber cycle is to find out __________
Answer:
lattice energy

Plus One Chemistry Thermodynamics Two Mark Questions and Answers

Question 1.
1. Explain enthalpy of fusion.
2. Give illustration of fusion of ice.
Answer:
1. It is the enthalpy change when one mole of a solid is converted into its liquid at its melting point.

2. Enthalpy of fusion of ice is 6 kJ/mol. From this it is clear that 6 kJ of energy is required to convert one mole of ice (18 g) into water at 0°C.

Question 2.
a) What do you meant by enthalpy of vapourisation?
b) Explain enthalpy of sublimation.
Answer:
a) It is the enthalpy change when one mole of a liquid is converted into its vapour at its boiling point.
b) It is the enthalpy change when one mole of a solid is converted into its vapour at its transition temperature.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
One equivalent of an acid reacts completely with one equivalent of a base in dilute solution.
1. Which type reaction is this?
2. HCl + NaOH → NaCl + H2O
On the basis of above equation, explain enthalpy of neutralisation.
Answer:
1. Nneutralisation.

2. When one equivalent of HCl (36.5 g) reacts completely with one equivalent of NaOH (40 g), 57.1 kJ energy is liberated.

Question 4.
1. What is the difference between system and surroundings?
2. There are different types of systems. What are they? Explain.
3. Give example for different types of systems.
Answer:
1. A system in thermodynamics refers to that part of universe in which observations are made. The remaining part of the universe other than the system constitutes the surroundings.

2. System is classified into the following three types. Open system: This is a system in which there is exchange of energy and matter between system and surroundings.
Closed system:
This is a system in which there is no exchange of matter, but exchange of energy is possible between system and the surroundings.

Isolated system:
This is a system in which there is no exchange of energy or matter between the • system and the surroundings.

3. Open system
Presence of reactants in an open beaker
Closed system
Presence of reactants in a closed vessel made of conducting material
Isolated system
Presence of reactants in a • thermos flask or any other closed insulated vessel

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 5.
Match the following:

A B
1. Isothermal Temperature varies
2. Adiabatic Temperature constant
3. Isobaric Volume constant
4. Isochoric Pressure constant

Answer:

A B
1. Isothermal Temperature constant
 2.Adiabatic Temperature varies
3. Isobaric Pressure constant
4. Isochoric Volume constant

Question 6.
1. What is meant by enthalpy?
2. Derive an equation for enthalpy change.
3. What is enthalpy change?
Answer:
1. Enthalpy is the sum of internal energy and pressure volume energy.
i.e.,H = U + pV

2. ∆H = ∆U + ∆pV)
∆H = ∆U + p∆V + V∆p
At constant pressure, ∆p=0
∆H = ∆U+p∆V
But ∆U=q+w
∆H=q+w+p∆V
w = -p∆V
i.e; ∆H= q – p∆V + p∆V
∆H = qp

3. Enthalpy change is heat absorbed or released at constant pressure.

Question 7.
1. Find the enthalpy of the reaction,
C(graphite) + O2(g) → CO2(g)
Given,
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =-110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?
Answer:
1. Considerthe reaction,
C(grahite) + O2(g) → CO2 (g); ∆H = x
CO2 can also be prepared through the following two steps:
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
Then by Hess’s law, x = (-110.5+-283.0) kJ=-393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.
For a spontaneous process ∆STotal is +ve.
For a nonspontaneous process ∆STotal is -ve.

Question 8.
Explain the following:

  1. Enthalpy of atomization
  2. Enthalpy of solution at infinite dilution

Answer:

  1. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
  2. It is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions or solute molecules are negligible.

Question 9.
The enthalpy change for the reaction,
N2(g) + 3H2(g) → 2NH3(g) is -92.38 kJ at 298 K.
What is ∆U at 298 K?
Answer:
∆U = ∆H — ∆ngRT
= -92.38 × 10³ J – [-2 × 8.314J K-1 mol-1 × 293 K)]
= -92.38 × 10³J +4.872 × 10³J
= -87.51 × 10³J
= – 87.51 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
What are the two types of heat capacities? How they are related?
Answer:
The two types of heat capacities are heat capacity at constant pressure (Cp) and heat capacity at constant volume (Cv). These two are related as Cp – Cv = R, where R is the universal gas constant.

Question 11.
Enthalpy and Entropy changes of two reactions are given below: Find out whether they are spontaneous or not at 27°C. Justify.
1. ∆H = 26 kJ/mole, ∆S = 8.3 J/K/mole
2. ∆H = -393.4 kJ/mole, ∆S = 6 J/K/mole
Answer:
1. ∆G = ∆H -T∆S
= 26000 – 300 × 8.3 = 23.510
Since ∆G is positive, the process is non-spontaneous.

2. ∆G = ∆H -T∆S
= -393400 – 300 × 6 = -391600
Since ∆G is negative, the process is spontaneous.

Question 12.
1. What is enthalpy of solution?
2. What is enthalpy of dilution?
Answer:
1. Enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure.

2. Enthalpy of dilution is the heat withdrawn from the surroundings when additional solvent is added to the solution. It is dependent on the original concentration of the solution and the amount of solvent added.

Question 13.
What is the significance of the second law of thermodynamics in the spontaneity of exothermic and endothermic reactions?
Answer:
The second law of thermodynamics provides explanation for the spontaneity of chemical reactions. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall 1 entropy change is positive which makes the reaction spontaneous.

In the case of endothermic reactions, since heat is ‘ absorbed by the system from the surroundings, the entropy change of the surroundings becomes negative (∆Ssurr < 0). In this case the process will be spontaneous only if the entropy change of the reacting system is postive (∆Ssys > 0) and is also greater than ASsurr in magnitude so that the overall entropy change (∆Stotal) is positive.

Question 14.
Explain the importance of third law of thermodynamics.
Answer:
The importance of third law of thermodynamics lies in the fact that it permits the calculation of absolute values of entropy of pure substances from thermal data alone. For a pure substance, this can be done by summing \(\frac { { q }_{ rev } }{ T } \) increments from 0 K to 298 K.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 1

Question 15.
C12H22O11 + 12O2 → 12CO2 + 11H2O
Consider this equation and answer the following questions.
a) Thermodynamically, which type reaction is this?
b) What is enthalpy of combustion?
c) Give another example.
Answer:
a) Combustion.
b) It is the enthalpy change when one mole of a substance undergoes complete combustion in excess of air or oxygen.
c) C6H12O6 + 6O2 → 6CO2 + 6H2O

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 16.
Bond dissociation energies of hydrogen and nitrogen are 430 kJ and 41.8 kJ respectively and the enthalpy of formation of NH3 is – 46 kJ. What is the bond energy of N-Hbond?
Answer:
3H2 + N2 → 2NH3; AH = -46 kJ
3 × ½H2 + ½N2 → 2 × ½NH3
[(3× ½ × 430) + (½ × 941 .8)] – (3N-H) = – 46
[(3 × 215) + (470.9) + (46)] → [3N-H]
[645 + 470.9 + 46] = 3N-H
N-H = 387.3 kJ

Plus One Chemistry Thermodynamics Three Mark Questions and Answers

Question 1.
In 1840, G.H.Hess (a Russian chemist) proposed an important generalisation of thermochemistry which is known after his name as Hess’s law.
1. State Hess’s law.
2. Give illustration of Hess’s law.
Answer:
1. Enthalpy change in a chemical reaction is same whether it takes place in one step or in more than one step.

2. Considerthe formation of CO2.
C + O2 → CO2; ∆H = x
CO2 can be prepared through the following two steps:
C + ½O2 → CO ; ∆H = y
CO + ½O2 → CO2; ∆H = Z
Then by Hess’s law,
x = y + z

Question 2.
∆U = q-p∆V. If the process is carried out at constant
volume, then ∆V=0. Answer the following questions.
1. Give the equation for ∆U.
2. 1000J was supplied to a system at constant volume. It resulted in the increase of temperature of the system from 45 °C to 50 °C. Calculate the change in internal energy.
Answer:
1. ∆U = qv

2. Since the volume kept constant, ∆V=0
∴ ∆U = qv = 1000J

Question 3.
Thermodynamics deals with macroscopic properties.
1. What is the difference between extensive and intensive properties?
2. Classify the following properties into extensive and intensive.
Pressure, Mass, Volume, Temperature, Density, Heat capacity, Viscosity, Surface tension, Internal • energy, Molar heat capacity, Refractive index, Enthalpy, Specific heat capacity
Answer:
1. Extensive properties are those properties whose value depend on the quantity or size of matter present in the system.
Intensive properties are those properties which do not dependent on the quantity or size of matter present in the system.

2. Extensive properties: Mass, Volume, Heat capacity, Internal energy, Enthalpy Intensive: Pressure, Temperature, Density, Viscosity, Surface tension, Molar heat capacity, Refractive index, Specific heat capacity.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 4.
1. What is meant by state of the system and state variables?
2. Give any four examples for state variables/state functions.
Answer:
1. The state of a system refers to the conditions of existence of a system when its macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. The measurable properties required to describe the state of a system are called state variables or state functions. A state function is a property of a system whose value depends only upon the initial and final states of the system and is independent of the path by which this state has been reached. Properties whose values depend on the path followed are called path functions.

2. State variables/State functions – Temperature, Pressure, Enthalpy, Entropy

Question 5.
1. Explain the Zeroth law of thermodynamics.
2. What are the important modes of transference of energy. Explain.
Answer:
1. The Zeroth law of thermodynamics states that if two bodies say, ‘A’ and ‘B’ are in thermal equilibrium with another body say, ‘C’, then the bodies A’ and ‘B’will also be in thermal equilibrium with each other. It provides the basis for the measurement of temperature.

2. The two important modes of transference of energy are heat and work.
Heat:
The exchange of energy, which is a result of temperature difference between system and surroundings is called heat (q).

Work:
The exchange of energy between system and surroundings can occur in the form of work which can be mechanical work, electrical work or pressure-volume work. The exchange of energy as pressure-volume work can occur if system consists of gaseous substance and there is a difference of pressure between system and surroundings.

Question 6.
1. Explain the symbols and sign conventions of heat and work.
2. Explain internal energy.
Answer:
1. Heat is represented by the symbol ‘q’. The ‘q’ is positive, when heat is transferred from the surroundings to the system and ‘q’ is negative when heat is transferred from system to the surroundings.
Work is represented by the symbol ‘w’. The ‘w’ is positive when work is done on the system and ‘w’ is negative when work is done by the system.

2. Every substance is associated with a definite amount of energy due to its physical and chemical constitution. This is called internal energy. It is the sum of the different types of energies such as chemical, electrical, mechanical etc.

Question 7.
Fill in the blanks.

  1. If heat is released, ‘q’ is ……………
  2. For exothermic process ‘∆H’is ………………
  3. If work is done on the system, ‘w’ is ………………
  4. For endothermic process ‘∆H’ is ………………
  5. If work is done by the system, ‘w’ is ………………

Answer:

  1. Negative
  2. Negative
  3. Positive
  4. Positive
  5. Negative

Question 8.
1. What is meant by enthalpy of formation?
2. What is the value of standard enthalpy of formation (∆fH) of an element?
Answer:
1. Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable states of aggregation (i.e., reference state).

2. Zeno

Question 9.
First Law of thermodynamics is the law of conservation of energy.

  1. Give the mathematical form of the first law.
  2. Write the Gibb’s equation.
  3. What is the sign for ∆G for a spontaneous process?

Answer:

  1. ∆U=q+w w = work done
    q = heat absorbed
  2. G = H- TS or ∆G= ∆H – T∆S
  3.  In the case of spontaneous process ∆G = -ve

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
1. Predict the sign of ∆S for the reaction,
NH3(g) + HCl(g) → NH4Cl(s)
2. The reaction between gaseous hydrogen and chlorine is
H2(g) + Cl2(g) → 2HCl(g); ∆rH = -1840 kJ
i) What is the enthalpy of formation of HCl?
ii) How much heat will be liberated at 298 K and 1 atm for the formation of 365 g of HCl?
Answer:
1. ∆S is negative

2. i) ∆fH = \(\frac{-1840}{2}\) = -920 kJ mol-1
ii) Heat liberated during the formation of 1 mole (36.5 g) of HCl = 920 kJ
∴ Heat liberated during the formation of 365 g of HCl = 9200 kJ

Question 11.
Derive the Meyer’s relationship.
Answer:
We have, q = C × ∆T
At constant volume, qv = Cv × ∆T = ∆U
At constant pressure, qp = Cp × ∆T = ∆H
For 1 mole of an ideal gas,
∆H = ∆U + ∆(pV) = ∆U + ∆(RT) = ∆U + R∆T
∴ ∆H = ∆U + R∆T
On putting the values of ∆H and ∆U,
Cp∆T = Cv∆T + R∆T
Cp = Cv + R
Cp – Cv = R, which is the Meyer’s relationship.

Question 12.
1. In a process 701J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
2. What is free expansion? What is the work done during free expansion of an ideal gas?
Answer:
1. ∆U=q+w = 9 + w = 701J – 394 J = 307J

2. Expansion of a gas in vacuum is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.

Question 13.
1. Name the instrument used for measuring the ∆U of a process.
2. What is the value of ∆G for a reaction at equilibrium?
3. ∆H and ∆S of a reaction are 30.56 and 0.666 kJ/ mol respectively at 1 atm pressure. Calculate the temperature at which the reaction is in equilibrium.
Answer:
1. Bomb calorimeter
2. Zeno
3. ∆H-T∆S = 0 or ∆H = T∆S
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 2

Question 14.
Thermodynamic process differ based on the manner
in which it is carried out.
1. Distinguish between reversible and irreversible processes.
2. Calculate the amount of work done when 2 moles of a gas expands from a volume of 2 L to 6 L isothermally and irreversibly against a constant external pressure of 1 atm.
Answer:
1.

Reversible process Irreversible process
1) Which can be reversed 1) Which cannot be reversed spontaneous process
2) Takes place infinitesimally slowly 2) Takes place spontaneous
3) Work done maximum 3) Work done minimum

2. w = -p∆V = -1 × (6 – 2)
= – 4 L atm

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 15.
1. What are thermochemical equations?
2. Give an example for a thermochemical equation.
Answer:
1. A balanced chem ical equation together with the value of its ∆rH is called a thermochemical equation.
2.

Question 16.
1. Define lattice enthalpy of an ionic compound.
2. What is Born-Haber cycle?
Answer:
1. The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

2. It is a simplified method developed by Max Born and Fritz Haberto correlate lattice enthanpies of ionic compounds to otherthermodynamic data.

Question 17.
Predict what happens to entropy in the following changes:

  1. Metal is converted into alloy.
  2. Solute crystallizes from solution.
  3. Hydrogen molecule dissociates.

Answer:

  1. The entropy will increase.
  2. The entropy will decrease.
  3. The entropy will increase.

Question 18.
1. Give the relation between change in enthalpy and change in free energy.
2. Name the above relation.
3. What is the significance of the above relation?
Answer:
1. ∆G = ∆H – T∆S

2. Gibbs equation orGibbs-Helmholtz equation.

3. This relation is used to predict the spontaneity of a process based on the value of ∆G . If ∆G is negatve, the process is spontaneous. If ∆G is positive, the process is non-spontaneous.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 19.
1. Predict in each of the following whether entropy increases or decreases.
i) Sublimation of camphor
ii) 4Fe(s) + 3O2(g) → 2Fe2O3(g)
2. The equilibrium constant for a reaction at 30 °C
2.5 x 10-29. What will be the value of ∆G?
Answer:
1. i) entropy increases
ii) entropy increases
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 4

Question 20.
1. Explain the effect of temperature on the spontaneity of a process based on Gibbs equation.
2. For a reaction 2A(g) + B(g) → 2D(g), enthalpy and entropy changes are – 20.5 kJ mol-1 and – 50.4 J K-1mol-1 respectively. Predict whether the reaction occurs at 25 °C.
Answer:
1. If ∆H is -ve and ∆S is +ve, ∆G would certainly be -ve and the process will be spontaneous at all temperatures.
If both ∆H and ∆S are – ve ∆G would be -ve if ∆H > T∆S
If both ∆H and ∆S are + ve ∆G would be -ve if T∆S > ∆H
If ∆H is +ve and AS is -ve, ∆G would certainly be +ve and the process will be non-spontaneous at all temperatures.

2. According to Gibbs equation, ∆G = ∆H – T∆S
∆G = (-20.5 × 10³)-(298 ×-50.4)
= – 20500 + 15019.2 = – 5480.8 J mol-1
Since ∆G is -ve, the process is spontaneous.

Plus One Chemistry Thermodynamics Four Mark Questions and Answers

Question 1.
1. Explain the first, second and third laws of thermodynamics.
2. What do you meant by entropy?
3. Explain the spontaneous process.
Answer:
1. First law:
Energy can neither be created nor be destroyed. Energy in one form can be converted into another form without any loss or gain.

Second law:
Entropy of the universe increases during a spontaneous process.

Third law:
Entropy of a perfect crystalline substance is zero at absolute zero of temperature.

2. Entropy:
Entropy is a measure of randomness or disorder of a system.

3. Spontaneous process:
A spontaneous process is defined as an irreversible process which has a natural tendency to occur either of its own or after proper initiation under the given set of conditions.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 2.
U1, q, w, U2 are given. U1 is internal energy, q is absorbed heat, w is work done and U2 is final energy.
a) Derive an equation for ∆U.
b) Give the equation for w.
c) Calculate the change in internal energy of a system which absorbs 200 J of heat and 315 J of work is done by the system.
Answer:
a) U2 = U1 + q +w
U2 – U1 = q + w
or ∆U = q + w
b) w= -p∆V
c) q = 200J
w = -315J
∆U = ?
∆U = q + w
= 200 + {315}
= 200-315 = -115J

Question 3.
a) Predict whether entropy increases or decreases in the following changes:
i) l2(s) → l2(g)
ii) Temperature of a crystalline solid is raised from 0 Kand 115 K.
iii) Freezing of water
b) Calculate the enthalpy of combustion of methane. Given that standard enthalpies of formation of CH4, CO2 and H2O are -75.2, -394 and -285.6 kJ/mol respectively.
Answer:
a) i) Entropy increases
ii) Entropy increases
iii) Entropy decreases
b) The required equation is,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 5

Plus One Chemistry Thermodynamics NCERT Questions and Answers

Question 1.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? (2)
Answer:
Heat absorbed by the system, (q) = + 701 J
Work done by the system (w) = – 304 J
Change in internal energy (∆U) = q + w
= 701 – 394
= 307 J

Question 2.
The reaction of cyanamide, NH2CN (s) with oxygen was carried out in a bomb calorimeter and AU was found to be – 742.7 kJ mol1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. (2)
Answer:
NH2CN(S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
∆U = 742.7 kJ mol-1;
∆n(g) =2 – 3/2= + 0.5
R = 8.314 × 10-3kJ K-1 mol-1;
T = 298K
According to the relation, ∆H = ∆U + ∆ngRT
∆H = -742.7 kJ + 0.5 mol × 8.314 × 10-3kJ K-1 mol-1 × 298 K
=-742.7 kJ + 1.239kJ
=-741.5 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol-1 K-1. (2)
Answer:
Moles of Al (n) = \(\frac { 60{ g } }{ 27{ g }{ mol }^{ -1 } } \) = 2.22 mol
Molar heat capacity (Cm) = 24 J mol-1 K-1
∆T = 55 °C – 35 C° = 20C° or 20 K
Now, q = Cm × n × ∆T
= 24.0 J mol-1 K-1 × 2.22 mol × 20 K
= 1065.6 J
= 1.067 kJ

Question 4.
The enthalpy of formation of CO(g), CO2 (g), N2O (g), N2O4 (g) are -110, – 393, 81 and 9.7 kJ mol-1 respectively. Find the value of ∆rH for the reaction:
N2O4 (g) + 3CO(g) → N2O(g) + 3CO2(g)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 6

Question 5.
The equilibrium constant for the reaction is 10. Calculate the value of ∆G ; Given R = 8 J K-1 mol-1; T = 300 K.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 7

Question 6
Calculate the entropy change in surroundings when 1.0 mol of HzO (I) is formed under standard conditions. Given ∆fH = – 286 kJ mol-1.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 8

Question 7.
Comment on the thermodynamic stability of NO(g) and NO2 (g) given :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 9
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 10

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Students can Download Chapter 9 Hydrogen Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Plus One Chemistry Hydrogen One Mark Questions and Answers

Question 1.
In which of the following compounds does hydrogen have an oxidation state of -1?
a) CH4 b) NH3 C) HCl d) CaH2
Answer:
d) CaH2

Question 2.
The radio active isotope of hydrogen is __________ .
Answer:
Tritium

Question 3.
Temporary hardness of water is due to the presence of
a) MgSO4
b)Ca(HCO3)2
c) CaSO4
d) NaHCO3
Answer:
b) Ca(HCO3)2

Question 4.
D2O is used as __________ in nuclear reactors.
Answer:
Moderator

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 5.
30 volumes of H2O2 means
a) 30% H2O2 solution
b) 30 cm³ of the solution contains 1 g of H2O2
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP
d) 30 cm³ of the solution contains 1 mole of H2O2
Answer:
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP

Question 6.
Name the three isotopes of hydrogen.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 1

Question 7.
Dihydrogen is prepared on industrial scale from syngas by
Answer:
Water gas shift reaction

Question 8.
Among the following elements which does not make a hydride is _________
a) Ti
b) Mg
c) Co
d) Pd
Answer:
c)Co

Question 9.
Hydrogen is purified by _________ .
Answer:
Oculusion on pd

Question 10.
H2O2 is _________ .
Answer:
Diamagnetic

Question 11.
Ortho & para hydrogen are __________ .
Answer:
Nuclear spin isomers

Plus One Chemistry Hydrogen Two Mark Questions and Answers

Question 1.
Write one method each for the laboratory preparation of dihydrogen from
i) mineral acid
ii) aqueous alkali.

  1. Which is the catalyst used for the reaction?
  2. Which is the product in this reaction?

Answer:
1.By the reaction of granulated zinc with dilute HCl.
Zn + 2HCl → ZnCl2 + H2

2. By the reaction of zinc with aqueous alkali.
Zn + 2NaOH → Na2ZnO2 + H2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 2.
Classify the following into those causing temporary hardness and permanent hardness:
[Mg(HCO3)2, MgCl2, CaCO3, CaSO4, NaCl, NaHCO3, Ca(HCO3)2]
Answer:
Temporary hardness – Mg (HCO3)2, Ca(HCO3)2
Permanent hardness – MgCl2, CaSO4

Question 3.
The bond angle in water is different from the tetrahedral bond angle.

  1. What is the bond angle and shape of water?
  2. Justify.

Answer:
1. 104.5°

2. There are three types of repulsions in water. They are: Ip – Ip repulsion, Ip – bp repulsion and bp – bp repulsion. In order to minimize the stronger Ip- Ip repulsion, the bond angle is reduced to 104.5° from 109.5°. Thus, the shape of water is distorted tetrahedral or angular.

Question 4.
Among NH3, H2O and HF which would you expect to have highest magnitude of hydrogen bonding? Why?
Answer:
Strength of H-bond depends upon the atomic size and electronegativity of the other atom to which H- atom is covalently bonded. Smaller size and higher electronegativity favour H-bonding. Among N, F and O atoms, F is the smallest and its electronegativity is highest. Hence, HF will have highest magnitude of H-bonding.

Question 5.
1. Soap does not give lather with hard water. Why?
2. What are the disadvantages of hard water?
Answer:
1. Hard water contains Ca+, Mg2+ions in the form of
their bicarbonates, chlorides or sulphates. Hard water forms scum/precipitate with soap. So soap does not give lather with hard water,

2. Hard water is unsuitable for laundry. It is harmful for boilers because of deposition of salts in the form of scale. This reduces the’ efficiency of the boiler.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Write one example each for the oxidising action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium, H2O2 oxidises PbS to PbSO4.
PbS(s) + 4H2O2(aq) → PbSO4(s) + 4H2O(l)
In basic medium, H2O2 oxidises Fe2+ to Fe3+.
2Fe2+ + H2O2 → 2Fe3+ + 2OH

Question 7.
Write one example each for the reducing action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium H2O2 reduces MnO4 to Mn2+.
2MnO4 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
In basic medium H2O2 reduces l2 to l.
l2 + H2O2 + 2OH → 2l + 2H2O + O2

Question 8.
Write two examples for redox reactions involving water.
Answer:
1. Water can be easily reduced to dihydrogen by highly electropositive metals like Na. Here, Na is oxidised to NaOH.
2H2O(I) + 2Na(s) → 2NaOH(aq) + H2(g)

2. With F2, water is oxidised to O2. Here, F2 is reduced to F-.
2F2(g) + 2H2O(l) → 4H+(aq) + 4F(aq) + O2(g)

Question 9.
Distinguish between

  1. Hard and Heavy water
  2. Temporary and permanent hardness of water.

Answer:
1. A sample of water said to be hard water if it does not give lather with soap. Heavy water is the oxide of deuterium (D2O).

2. Temporary hardness is due to the presence of bicarbonate of calcium or magnesium. It can be removed by boiling.

Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium. It is not removed by boiling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 10.
Hydrogen combines with elements to give binary compounds known as hydrides.
1. Name the three categories of hydrides.
2. Classify the given hydrides into different categories: NH3 LiH, TiH, CH4, NaH
Answer:
1. Ionic hydrides, Covalent hydrides and Metallic/ Interstitial hydrides

2. NaH, LiH – Ionic hydrides
NH3, CH4 – Covalent hydrides
TiH – Interstitial hydride

Plus One Chemistry Hydrogen Three Mark Questions and Answers

Question 1.
A chart prepared by a student based on the similarities of hydrogen with alkali metals and halogens is as shown below. Correct the mistakes in it.

Similarities with alkali metals Similarities with halogen
-1 oxidation state Reducing agent + 1 oxidation state High Ionisation energy
Diatomic state During electrolysis, both of them are produced at the cathode
Non metals

Answer:

Similarities with alkali metals Similarities with halogen
+1 oxidation state Reducing agent -1 oxidation state High Ionisation energy
During electrolysis, both of them are produced at the cathode Diatomic state
Non metals

Question 2.
Prepare a short note on different types of hydrides.
Answer:
Hydrogen reacts with metals or non metals to form binary compounds known as hydrides.
Hydrides are mainly classified into the following 3 types:
1. Ionic or salt like hydrides
These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. These are crystalline, non-volatile and non-conducting in solid state. But their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode. They react violently with water producing dihydrogen gas. e.g. NaH, KH.

2. Covalent hydrides or Molecular hydrides
These are formed by the action between dihydrogen and non metals (p-block elements). Covalent hydrides are classified into electron-deficient (e.g. B2H6), electron-precise (e.g. CH4) and electron-rich hydrides (e.g. NH3).

3. Metallic or Non-stoichiometric or Interstitial hydrides
These are formed by the reaction of dihydrogen with many d-block and f-block elements. These hydrides conduct heat and electricity. They are almost always non- stoichiometric, being deficient in hydrogen, e.g.
LaH2.87,YbH2.55,VH0.56 etc.

Question 3.
Consider the chemical equation and fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 2

  1. This method is used for the preparation of ……………
  2. The electrode used is …………
  3.  ……….. is produced at cathode.

Answer:

  1. Hydrogen
  2. Platinum
  3. Hydrogen

Question 4.
Analyse the equation: 2Na + H2 → 2NaH

  1. In this chemical equation, H2 reacts with …………
  2. Hydrogen reacts with metals to form …………
  3. Hydrogen is in ………… oxidation state in NaH.

Answer:

  1. Na
  2. Metal hydrides
  3. -1

Question 5.
a) How is hydrogen of high purity prepared?
b) Dihydrogen is relatively inert at room temperature. Give reason.
c) Write any two uses of hydrogen.
Answer:
a) High purity (>99.95%) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide . solution bewteen nickel electrodes.
b) This is due to high H-H bond enthalpy.
c) 1. In oxy-hydrogen torches.
2. Liquid hydrogen is used as a fuel in rockets.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
A sample of cold river water does not easily give lather with soap, but on boiling it does.

  1. Evaluate and write the chemical equation involved.
  2. In some cases, the water does not give ready lather even if it is boiled. Why?

Answer:
1. If the sample of water has temporary hardness, it is removed during boiling. Here, the bicarbonates of magnesium is precipitated as Mg(OH)2 and that of calcium is precipitated as CaCO3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 3

2. The sample of water might have permanent hardness due to the presence of dissolved chlorides and sulphates of Ca and Mg. This cannot be removed just by boiling.

Question 7.
Hydrogen has three isotopes – Protium, Deuterium and Tritium.
a) Of these which is the radio active one?
b) Name a compound which contains the isotope deuterium.
c) Make a table which shows number of protons, neutrons and electrons in each isotope.
Answer:
a) Tritium
b) Heavy water – D2O
c)

Isotope No. of Protons No. of Neutrons No. of Electrons
Protium 1 0 1
Deuterium 1 1 1
Tritium 1 2 1

Question 8.
What are the three types of hydrates? Give examples for each.
Answer:
1. Hydrates with coordinated water, e.g. [Cr(H2O)6]Cl3.
2. Hydrates with interstitial water, e.g. BaCl2.2H2O
3. Hydrates with hydrogen bonded water, e.g. CuSO4.5H2O.

Question 9.
1. What is meant by amphoteric nature of water?
2. Suggest an example to show this property.
Answer:
1. The amphoteric nature of water means that it has the ability to act as an acid as well as a base.

2. Water acts as an acid with NH3 and a base with H2S.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 4

Question 10.
a) What are the advantages of dihydrogen as a fuel?
b) What are the disadvantages of using dihydrogen as a fuel?
c) What is meant by the term ‘hydrogen economy’?
Answer:
a) 1. It is abundantly available in the combined state as water.
2. Use of dihydrogen as fuel provides pollution free atmosphere because its combustion product is only water.
3. Heat of combustion per gram of dihydrogen is more than twice that of jet fuel.

b) 1. Dihydrogen does not occur in free state in nature.
2. Hydrogen gas has explosive flammability which causes problem to its storage and transportation.
3. A cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol containing the same amount of energy.

c) Hydrogen economy refers to the use of dihydrogen as an alternative source of energy. The basic principle of hydrogen economy is storage and transportation of energy in the form of dihydrogen instead of fossil fuels or electric power.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 11.
What do you mean by temporary and permanent hardness? Explain one chemical method each for removing temporary hardness and permanent hardness.
Answer:
Temporary hardness is due to the presence of bicarbonates of calcium or magnesium. Permanent hardness is due to the presence of chlorides and sulphates of calcium or magnesium ions.
Clark’s method :
This method is used for removing temporary hardness. In this method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 5
Treatment with washing soda: This method is used to remove permanent hardness. Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.

Question 12.
a) Write a method for the preparation of H2O2.
b) What is 100 volume H2O2?
c) Draw the structure of H2O2 in gas phase.
Answer:
a) H2O2 is prepared by acidifying barium peroxide and removing excess water by evaporation under reduced pressure.
BaO2.8H2O(s) + H2SO2(aq) → BaSO4(s) + H2O2(aq) + 8H2O(I)

b) A 30% solution of H2O2 is called 100 volume H2O2. It means that one mL of 30% H2O2 solution will give 100 V of oxygen at STP.

c) H2O2 is has non-planar structure as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 7

Question 13.
lon exchange process is commonly employed for large scale production of soft water.

  1. What is the basic principle involved in ion exchange method of softening water?
  2. What are inorganic cation exchangers? Give example.

Answer:
1. Adsorption

2. These are complex inorganic salts like sodium- aluminium silicate NaAlSiO4 which can exchange cations such as Ca2+ and Mg2+ ions in hard water for Na+ ions. e.g. Zeolite.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 14.
1. What is heavy water? Mention one use of heavy water.
2. Explain why hydrogen peroxide is not stored in glass vessels.
3. What is calgon? What is its use?
Answer:
1. D2O. Used as moderator in nuclear reactors.

2. To prevent its decomposition.

3. Calgon is chemically sodium hexametaphosphate (Na6P6O18). It is used to remove Ca2+ and Mg2+ ions of hard water by converting them into soluble complexes.
M2+ + Na4P6O182- → [Na2MP6O18]2- + 2Na+ (M = Mg,Ca)

Question 15.
1. Name the oxide of isotope of hydrogen used in nuclear reactor.
2. What are cation exchange resins? What is their role in removing permanent hardness of water?
Answer:
1. Heavy water (D2O).

2. Cation exchange resins contain large organic molecule with -SO3H group and are water insoluble. Ion exchange resin (RSO3H) is changed to RNa by treating with NaCl. The resin exchanges Na+ ions with Ca2+ and Mg2+ ions present in hard water to make the water soft.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 16.
1. What is permuitit? How is it useful in removing permanent hardness of water?
2. Compare the structures of water and hydrogen peroxide.
Answer:
1. Hydrated sodium aluminium silicate is called permutit (NaZ). When it is added to hard water it exchanges Ca2+ and Mg2+ ions with Na+ ions.
2NaZ(s) + M2+(aq) —> MZ2(s) + 2Na+(aq) M = Ma, Ca)

2. Water molecule has an angular or bent shape. H2O2 has a non-planar structure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 8

Question 17.
1. Give the industrial method of preparation of H2O2.
2. How is heavy water prepared?
3. How is pure de-mineralised water obtained?
Answer:
1. Industrially H2O2 is prepared by the auto-oxidation of 2-ethylanthraquinol.

2. Heavy water is prepared by exhaustive electrolysis of water or as a byproduct in some fertilizer industries.

3. Pure de-mineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In the cation exchange process, the H+ exchanges for Na+, Ca2+, Mg2+ and other cations present in water. This process results in proton release and makes the water acidic.
2RH(s) + M2+(aq) \(\rightleftharpoons \) MR2(s) + 2H+(aq)

In the anion exchange process, the OH exchanges for anions like Cl, HCO3, SO42- etc. present in water.
RNH3+OH(S) + X(aq) \(\rightleftharpoons \) RNH3+X(s) + OH(aq)

The OH ions, thus liberated neutralise the H+ ions set free in the cation exchange process to get pure de-mineralised water.
H+(aq) + OH(aq) → H2O(l)

Question 18.
1. Explain why hydrogen peroxide is stored in coloured plastic bottles.
2. Write any two uses of H2O2.
Answer:
1. In the presence of metal surfaces or traces of alkali (present in glass conatiners), the decomposition of H2O2 (2H2O2 → 2H2O + O2) is catalysed. It is therefore stored in wax-lined glasses or plastic vessels in dark.

2. 1. As a hair bleach and as a mild disinfectant.
2. In the syntheseis of hydroquinone, tartaric acid and certain food products and pharmaceuticals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 19.
1. A sample of hard water was found to loose its hardness on boiling. Name the type of hardness associated with this sample. 0
2. Write the name and formula of four minerals which cause permanent hardness of water.
3. What are electron deficient hydrides? Whether they behave as Lewis acids or Lewis bases? Why?
Answer:
1. Temporary hardness

2. CaCl2, MgCl2, CaSO4, MgSO4

3. These are covalent hydrides having too few electrons for writing conventional Lewis structure. They act as Lewis acids because they can accept electrons, e.g. B2H6

Plus One Chemistry Hydrogen Four Mark Questions and Answers

Question 1.
Match the following:

A B
Temporary hardness p-block elements
Hydrides Reducing agent
Permanent hardness Chloride
Alkali metals Bicarbonate

Answer:

A B
Temporary hardness Bicarbonate
Hydrides p-block elements
Permanent hardness Chloride
Alkali metals Reducing agent

Question 2.
Certain samples of water do no produce easy lather with soap.

  1. What is this condition of water called?
  2. Which are two types of this condition?
  3. Suggest two methods to change this condition of water.
  4. What are the problems caused by this condition of water?

Answer:

  1. Hardwater.
  2. Temporary hardness and permanent hardness.
  3. By boiling water and by adding Na2CO3
  4. Wastage of soap and boiler explosion.

Question 3.
Match the following:

1. D2O a) Hendry Cavendish
2. Hydrogen b) Water gas
3. Ca + H2 c) 31H
4. C2H4 + H2 d) Heavy water
5. Tritium e) 21H
6. Deuterium f) CaH2
7. CO + H2 g) C2H6
8. CO + Z2 h) Producer gas

Answer:
1) – d)
2) – a)
3) – f)
4) – g)
5) – c)
6) – e)
7) – b)
8) – h)

Question 4.
1. A list of compounds are given below:
H2O, HCl, CH4
Construct chemical reactions to show the preparation of H2 from each of the above compounds.
2. What is syn gas? How is it prepared?
3. What is coal gasification?
4. Explain water gas shift reaction.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 9
2. Syn gas is a mixture of CO and H2. It is prepared by passing steam over red hot coke.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 10

3. The process of production of syngas from coal is called coal gasification.

4. The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst. This is called water gas shift reaction.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 11

The CO2 is removed by scrubbing with sodium arsenite solution.

Plus One Chemistry Hydrogen NCERT Questions and Answers

Question 1.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? (2)
Answer:
The various isotopes of hydrogen are:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 12

Question 2.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions? (2)
Answer:
Hydrogen atom has only one electron and thus, to achieve stable inert gas configuration of helium, it shares its single electron with electron of another hydrogen atom to form a stable diatomic molecule. The stability of H2 is further confirmed by the fact, that formation of one mole of gaseous H2 molecules results in the release of 435.8 kJ of energy.
H(g) + H(g) H2(g); ∆H = – 435.8 kJ mol-1

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 3.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain. (2)
Answer:
In some of the transition metal hydrides, hydrogen is absorbed as H-atoms. Due to the inclusion of H- atoms, the metal lattice expands and thus becomes less stable. Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely divided metal. The hydrogen evolved in this manner can be used as a fuel. Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

Question 4.
What causes the temporary and permanent hardness of water? (2)
Answer:
Temporary hardness is caused by presence of bicarbonates of calcium and magnesium, i.e., Ca(HCO3)2 and Mg(HCO3)2 in water whereas permanent hardness is caused by presence of soluble chlorides and sulphates of calcium and magnesium, i.e., CaCl2, CaSO4, MgCl2 and MgSO4 in water.

Question 5.
Write chemical reactions to show amphoteric nature of water. (2)
Answer:
Water is amphoteric in character. It means that it can act as proton donor as well as proton acceptor. When it reacts with acids, (stronger than itself), it behaves as a base. When it reacts with bases (stronger than itself) it acts as an acid.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? (2)
Answer:
Demineralised or distilled water is not useful for drinking purposes because it does not contain even useful minerals. Therefore, to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised or distilled water.

Question 7.
How does H2O2 behave as a bleaching agent? (2)
Answer:
The bleaching action of H2O2 is due to the nascent oxygen which is liberates on decomposition.
H2O2 → H2O + [O]
The nascent oxygen oxidise the colouring matter to colourless products. Hence, H2O2 is used for the bleaching of delicate maerials like ivory, feather, silk, wool, etc.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

You can Download Let’s Regain Our Fields Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

Let’s Regain our Fields Questions and Answers

Food Safety

Food Safety is a condition in which every individual are provided with food enough to lead a healthy life. It is very’ essential to ensure food safety to create a society in which there is no fear of poverty and health issues due to malnutrition. It is a challenge to ensure food safety in a condition in which the agricultural fields are declining. A culture that loves soil and agriculture has to be recreated. It is also important to regain the lost agricultural fields.

Indicators (Text Book Page No:35)

Question 1.
Didn’t you notice the illustration and the newspaper report?
Answer:
Population explosion, insufficient agricultural fields, negative attitude towards agriculture, unscientific vision, methods of agriculture, influence of consumer society, etc., are the causes of scarcity of food.

Let’s Regain our Fields Question 2.
What is the concept indicated by the illustration?
Answer:
Modern techniques of agriculture play a major role to overcome the problems and crises mentioned. The techniques agriculture like polyhouse farming, open precision farming, integrated pest management, hydroponics. aeroponics etc are the contributions of science. The novel information and findings form remedy for food scarcity.

HSSLive.Guru

8th Class Science Question 3.
Discuss it with your friends using the given indicators. Write your inferences in the science diary.
Answer:
The major goal of food safety is to eliminate hunger and poverty from our country. Supplying food grains in cheaper rates is ultimately beneficial to the people of the lower strata.

Indicators (Text Book Page No:36)

Question 4.
What were the changes that occurred in the area of agricultural fields from the year 1971 to 2011?
Answer:
During the period 1971-2011, the area of agricultural fields considerably reduced. In 1971 the area was 8.75 lakh hectors and in 1991 it was reduced to 5.5 lakh hectors. It was reduced to 2.08 lakh hectors in the year 2011.

Precision Pest Control Question 5.
What tendency could be observed in rice production and population growth during the period?
Answer:
In this period population increased from 2.13 crores to 3.34 crores. But the production of rice steeply reduced from 13.65 lakh tons to 5.69 lakh tons.

Polyhouse Sheet Price in Kerala Question 6.
Is this tendency desirable? Why?
Answer:
This tendency is not advisable. The increase in population and decrease in the area of crop field and the production of rice may lead the nation to poverty. The goal of food safety will not be attainable.

Question 7.
What are the obstacles faced by farmers today?
Basic Science Question and Answer:
The Crises in the field of Agriculture

Farmers face a number of crises. They are mainly loss, cost of production, limited cultivable land, climate change, exploitation of brokers, environmental destruction and health issues, etc. Many of these problems can be overcome if a positive attitude towards agriculture develop. There are many possibilities to solve every problems.

i. Fertile soil

Around 20 different elements are required for the growth of plants. They are called essential elements.
eg: Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, Potassium, Sulphur

These elements are naturally available in the soil by the decomposition .of microorganisms. The fertility of soil can be increased by proper manuring. Organisms like bacteria, fungi, algae, termite, earthworm, etc., help to increase the fertility of soil.

Indicators (Text Book Page No:38)

Question 8.
What is the role of microorganisms in ensuring the natural availability of elements in the soil?
Answer:
The elements are naturally available in the soil by the decaying action of micro-organisms. When leguminous plants are grown in fields, the microorganisms that harbour in the root nodules fixes atmospheric nitrogen to the soil, bacteria, fungi, algae, termite, earthworm, etc., increase the fertility of the soil.

HSSLive.Guru

8th Standard State Syllabus Question 9.
What is the need of testing the soil?
Answer:
Soil analysis (soil testing) is done in order to detect the amount of elements present and their pH value of the soil.

Question 10.
Why does the application of fertilizers become essential for better crop yield?
Answer:
Fertility of the soil is increased by proper manuring. Manuring helps to make up the deficiency of elements in the soil that are required for plant growth. As a result better yield is obtained.

Microbial Fertilizer

Microbial fertilizers are substances that contain micro-organisms that facilitate the fertility of the soil. Micro-organisms increase the amount factors, required for plant growth, in the soil. The bacteria like rhizobium, Azotobacter, azospirillum, and the aquatic plant azolla increase the amount of nitrogen in soil. Many things are to be taken care to retain the microorganisms in the soil. They are

  • Ensure the availability of organic fertilizer(biofertilizers) in the soil.
  • Sufficient water supply must be there.
  • Do not use chemical fertilizers and pesticides.

Consequences of Unscientific application of fertilizers

Indicators (Text Book Page No:39)

Question 11.
What are the consequences of unscientific application of chemical fertilizers? Discuss on the basis of the following indicators.

  • composition of soil
  • microorganisms in soil
  • health issues
  • financial factors

Answer:
Unscientific fertigation loses the fertility and changes the natural texture of soil. Certain elements required for plant growth will exceed in the soil and certain others become insufficient. Chemical fertilizers do not provide all the elements required for plant growth in proper amount and proportion.

Excessive fertigation and the use of pesticides kill the microorganisms in the soil. Thus the natural fertility of the soil loses.

The chemical substances present in chemical fertilizers get accumulated in agricultural crops (through biological magnification). It causes many chronic diseases in organisms including man who consume this food.

Chemical fertilizers and pesticides are expensive. Fertilizers are manufactured by small scale industries as well, as multinational companies. The expenses is not affordable by the farmers. Thus unscientific application of fertilizers causes consequences of varied dimensions.

Pest Control

Many methods are adapted today to control pests. Chemical pesticides are widely used. Chemical pesticides kill the pest as a whole. Use of chemical fertilizers cause a number environmental and health problems. High amount of chemical fertilizers are reported in ground water too.
Another possibility is to control pests using ultrasonic sound waves.
Using radiation the reproductive capacity of male pests can be lost and makes pest control effective.
Pheromone traps like devices also makes pest control effective. Pest control using the natural enemies of pest is highly effective and having no environment and health consequences.

Natural Enemies of Pest (Friendly Pests)

Organisms that eat pests, cause disease to them or parasitises on them are called their natural enemies.

Integrated Pest Management

In this method, the use of chemical pesticides is highly reduced. Pest control is made possible by the combined use of biological pesticides, friendly pests, mechanical control methods, etc. It will not disturb the equilibrium of the ecosystem.
It do not destroy pests as a whole. But it prevents the multiplication of pests and the number of pest is cortrolled to prevent crop loss.

Advantages of Integrated Pest Management

  • Do not kill the pests as a whole
  • Use of chemical pesticides is highly reduced.
  • No environmental or health issues as friendly pests, mechanical control measures, Biological pesticides, etc. are used.
  • Does not disturb the equilibrium of ecosystem.

Waste Management

Indicators (Text Book Page No:41)

Question 12.
Waste management and sustainable agriculture
Answer:
Live Stock management, Poultry farming, Pisciculture etc, help not only to earn income but also to the treatment of biological wastes. Composting, biogas production etc, are possible by using bio wastes. Cow dung is a very good biological manure and an essential component . for the production of biogas. By preparing cattle feed, fish food, poultry feed, etc., from bio waste more earnings can be done and biological waste management also possible.
Let's Regain our Fields in Malayalam

Certain methods of Agriculture

Many farming techniques that help to earn improved income by scientific approach are in practice.
eg: Rearing of cattle’s, Poultry farming, Sericulture, Pisciculture, Floriculture, Apiculture, Cuniculture, Mushroom culture, Horticulture, Medicinal plant cuitivation etc.
→ Live stock Management
Rearing of cattle’s for milk, meat and agricultural purposes.
→ Poultry farming
Rearing of birds for egg, meat etc.
→ Sericulture
Rearing of silk worms
→ Pisciculture
Rearing of fishes
→ Floriculture
Cultivation of flowers for commercial purposes.
→ Apiculture
Rearing of honeybees
→ Cuniculture
Rearing of rabbits
→ Mushroom culture
Cultivation of mushrooms .
→ Horti culture
Cultivation of fruits and vegetables.

Completion of Table(Text Book Page No:44)

Njodiyan Honey Bee Question 13.
Complete the following table related to various agricultural sectors.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields 2

Answer:

Areas Products Varieties
Pisciculture fish, fish liver oil Pearl spot, Rohu
Apiculture Honey, Wax Kolan, Mellifera Njodiyan
Mushroom culture Edible mushroom Milk mushroom, Button mushroom
Livestock management Milk, Meat jercy(Cow) Murrah (Buffalo) Jamnapari (Goat)
Cuniculture Meat Grey giant, White giant
Sericulture Silk thread Mulberry silkworm Tussar silkworm Muga silkworm
Poultry farming Egg, Meat Athulyaf(hen) Muscovy (duck) Bobwhite (Quail)

Modern Techniques of Agriculture
1. Polyhouse farming
Polyhouse is a special arrangement made by completely or partially covering transparent sheet like polythene. Humidity and temperature kept constant in polyhouse. So plant growth will be fast. Nutrients are dissolved in water and sprayed. Pest attack also will be less.

2. Open Precision farming
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil, etc., are studied accurately with the help of modern technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

3. Hydroponics
Plants are grown in nutrient solution.
4. Aeroponics
Plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Indicators (Text Book Page No:46)

Question 14.
How are modern agricultural practices helpful in reducing crop loss due to climate change?
Answer:
In arrangements like polyhouses temperature and humidity are maintained constant. Plant grows fast and gets better yield. Polyhouse farming helps to reduce crop less due to climatic changes.

HSSLive.Guru

Question 15.
What are the advantages of precision farming?
Answer:
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil etc., are studied accurately with the help of modem technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

Question 16.
How does cultivation become possible without depending on soil?
Answer:
• Hydroponics and aeroponics are soilless cultivation methods.
• Plants are grown in nutrient solution.
• Aeroponic plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Farmers groups

Nowadays farmers groups, that provide farmers an opportunity to sell and buy their products without brokers mediators are very lively. Online gathering of farmers also is widespread as the demand for organic product raised. These online groups help to find customers for the quality organic products and to get good price for the products.

Completion of Table(Text Book Page No:51)

Question 17.
Complete the following table, adding important ideas.
Njodiyan

Answer:

Challenges Remedies
Climatic change • Polyhouse farming
• Hydroponics
Environmental distruction and health hazards • Scientific manuring
• Integrated Pest manage ment. Biological waste Management
Cost of production • Microbial fertilizers
• Integrated farming
Crop loss • Polyhouse farming
• Open precision  farming
• Integrated Pest management
Limited space • Hydroponics
• Aeroponics
• Terrace farming Loss
Loss • Farmers societies

• Online gatherings

Let’s Regain Our Fields Let us Assess (Text Book Page No:52)

Question 1.
Cuniculture is related to
a. Keeping of honey bees
b. Rearing of rabbits
c. Cultivation of fruits and vegetables
d. Rearing of fish
Answer:
Rearing of Rabbits

Question 2.
High quality hybrid varieties provide high yield. Then, what is the need of native varieties? Record your response to this statement.
Answer:
The native varieties exist in a particular locality by acquiring natural resistance and adaptations to the climate, availability of food, nature of soil etc., though they have low productivity.
The native varieties have high resis-tance to diseases and environmental conditions. The extinction of native varieties leads to the depletion of our biodiversity. New varieties can be developed only from the native varieties.

 

Question 3.
Which is the most appropriate way to reduce crop loss due to pests?
a. Using high concentration pesticides
b. Protecting friendly pests.
c. Practicing integrated pest management
d. Applying organic pesticides only.
Answer:
C  Adopt integrated Pest control

Question 4.
‘Lower price during higher yield’. Suggest a practical solution to overcome this crisis faced by farmers.
Answer:
Collect and distribute resources through farmers societies.

Let’s Regain Our Fields Additional Questions and Answers

Question 1.
Find the odd one in each group.
Also write the common characterestic of the others.
a. Compost, Microbial fertilizer, urea, Bone-meal
b. Anthurium, Sida, Ramacham, Koovalam
c. Boer, Litchi, Rambutan, Durian
d. Phosphorus, Potassium, Nitrogen, Azetohactor
e. Rock Phosphate, Factompho- se, Muriate of Potash, Malatheon,
f. Kuthiravaly, Jyothy, Thriveni, Jaya
g. Fowl Cholera, Anthrax, Chores, Ranikatt
h. Blight disease, Root wilt, Bud rot
i. Ammonia, Urea, Compost, Factomphose.
j. APIS (njodiyan); Naaran, Melliferra.
k. Minorka, Royans, Ankona, Gramalakshmi.
Answer:
a.Urea – This is a chemical fertilizer. Others are Bio-fertilizers.
b. Anthurium – This is an ornamental plant. Others are medicinal plants.
c. Boer – It is a variety of goat. Others are fruit varieties.
d. Azetohactor – Others are essential elements.
e. Malatheon – This is a chemical pes ticide others are chemical fertilizers.
f. Kuthiravaly – This a hybrid vari ety pepper plant others are vari eties of paddy.
g. Anthrax : This is a cattle disease others are fowl diseases.
h. Blight disease, It is affect on rice plant others are diseases of coconut tree.
i. Compost – This is an organic fertiliser, others are chemical fertilisers
j. Naaran. This is prawn.Others are Honey bees used in apiculture.
k. Royans is good quality ducks, Others are fowl varieties.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Rearing of honey bees : Apiculture; Rearing of silkworms : …………..
b. Malabari : Goat; Vechoor : …………..
c. Cowdung: ………….. ; Azospirillum : Microbial Fertilizer
d. Jersey: Cow; Murrah: …………..
e. Mushroom culture : Growing mushroom; Cuniculture : …………..
f. Malatheon : Chemical pesticide; Neem seed kemal: …………..
g. ………….. – Virus; Anthrax Bacteria.
h. Native variety : vechoor; Hybrid variety : …………..
i. Paddy- ………….. ; Arecanut plant : Mahali
j. Cocount caterpillar – ………….. ; Gamboosia- Larvae of mosqitoes
k. …………..: Domestic hen; Pekkins : Domestic Duck.
l. Blight disease- Paddy; Quick wilt- …………..
m. Organic fertiliser – Vermiw- ash; ………….. – Factomphose.
Answer:
a. Sericulture
b. Cow
c. Organic fertilizer
d. Buffalo
e. Rabbit farming
f. Organic pesticide
g. Foot and Mouth disease
h. Jersey cross
i. Blast disease
j. Ichneumon wasp
k. Royans is good quality ducks, Others are fowl varieties.
l. pepper.
m. chemical fertilisers

HSSLive.Guru

Question 3.
Arrange the following items from column B ,C with column A.

A
Agricultural sectors
B
Speciality
C
Varieties
a. Horticulture i. scientific way of rea­ring of honey 1. Ankora
b. Apiculture ii. scientific way of rearing of rabbits 2. Litchi
c. Cuniculture iii. scientific way of rearing silkworm 3. Njodiyan
iv. scientific cultiva­tion of fruits and 4. Naran

Answer:

A
Agricultural sectors
B
Speciality
C
Varieties
Horticulture scientific way of rea­ring of honey Litchi
Apiculture scientific way of rearing of rabbits Njodiyan
Cuniculture scientific way of rearing silkworm Ankora

Question 4.
Group the following statements into suitable for Polyhouse farming and Precision farming.
a. limiting the irrigation by covering soil with polyethene sheet.
b. Cover the field completely or partially by transparent polyethene sheet.
c. By regulating temperature and moisture constantly.
d. Selecting appropriate crop for agriculture only jifter understanding characters of soil, amount of elements present in soil and the presence of water.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields 4
Answer:

  • polyhouse farming – (b), (c)
  • Precision farming – (a), (d)

Question 5.
“It is essential to retain indegenious species of mango tree like Muvandan, Kilichundan even many hybrid varieties are avail-able”. Write your opinion to the farmer’s statement? Justify your answer?
Answer:

  • Farmer’s opinion is right.
  • Indigenous varieties of a locality are varieties that acquire natural immunity by adapting to the climate, the availability of food, soil texture of the place etc.
  • We can develop new high quality varieties only from indigenous varieties.

Question 6.
Find suitable term and fill the blanks;
Moovandan – Mango Tree Kasaragodu Dwaf – a. …………..
Njalippoovan – Musa Attappadi black – b. …………….
Answer:
a. Cow
b. Goat

Question 7.
Sustainable farming is an environment friendly method. Explain the reason.
Answer:
The excessive use of chemical fertilizers and pesticides may give increased profit. But this will not last long. The continuous use may spoil the natural fertility of the soil and the farmland may be changed into a barren land. By integrated cropping method, the use of outside manures, pesticides etc can be reduced. The wastes of one can be used as a manure or food for some other one. This will help to maintain the natural fertility of the soil. Moreover, the biodiversity also is conserved.

Question 8.
The practice of cultivating fruits and vegetables.
Answer:
Horticulture

HSSLive.Guru

Question 9.
Some statements regarding modern agricultural practices are given below. Which agricultural practice is related to this?
a. The method of farming in which nutrients are dissolved in water and are supplied on plants through dip irrigation.
b. The method plants are grown in nutrients solution
c. The method of farming by covering the rool using polythene sheets and by limiting.
Answer:
a. Polyhouse farming
b. hydroponics
c. precision farming

Question 10.
Polyhouse farming will be advantageous only in farmlands where cultivation is continuously maintained. Why?
Answer:
The cost of making a polyhouse is very high. But the yield from the crops will increase substantially.

Question 11.
Which of the following is not desirable in integrated pest control method?
a. Mechanical Pest Control
b. Excessive use of chemical pesticide
c. Friendly pests
d. Use of biopesticides
Answer:
b. Excessive use of chemical pesticide

Question 12.
What is the difference between polyhouse farming and open precision farming
Answer:
Poly house farming

  • Agricultural land is partially or completely covered by transparent polythene sheet.
  • Heat and humidity are kept constant
  • Nutrients are dissolved in water and given to plants through drip fertigation. Pest attack is compara-tively low.

Open Precision farming

  • Soil is covered by polythene sheet
  • Nature of soil, amount of elements in soil, pH of soil, presence of water etc are studied accurately using modern technology
  • Limited irrigation is needed.
  • Weed control is effective

Question 13.
What are the main characteristics of hybrid varieties?
Answer:
High yield, disease resisting capacity ability to give high yield within a short period etc. are the characteristics of hybrid varieties.

HSSLive.Guru

Question 14.
What are the consequences of unscientific application of chemical fertilizers?
Answer:

  • Financial loss.
  • Chemical pollution.
  • Destruction of microorganisms
  • Health issues.

Question 15.
How can grow plants without soil?
Answer:
Science has proved that cultivation is possible in the absence of soil, for example aeroponics and hydroponics. In hydroponics, plants are grow in nutrients solution. In aeroponics, plants are grow in such a way that their roots grow into air and nutrients are sprayed directly on roots.

Question 16.
What are the different varieties of buffaloes?
Answer:
Bhadawari, Jaffrabadi, Marwari, pashmina, malabari, Beetance.

Question 17.
Find out the odd one. Write the common feature of others. Rhizobium, Azetobacter, Lacto bacillus, Azospirillum
Answer:
Lactobacillus – others are nitrogen-fixing bacteria.

Question 18.
Arrange the organisms properly in the given table. Names are mentioned in the box.
Cadopt Technologies
Answer:
a. Jersey, Murrah
b. Athulya, Muscovy
c. Naran, Kara
d. Grey giant, Ankora
e. Muga, Tusser
f. Mellifera, Kolan

Question 19.
What is the difference between hydroponics and aeroponics?
Answer:

  • Hydroponics is the technique by which plants are grown in nutrient solution
  • In aeroponics plants are grown in such a way that their roots are penetrating towards air and nutrients are directly sprayed to their roots.

Question 20.
What are the advantages of precision farming?
Answer:
In this method of farming the nature of soil, quality of elements in the soil, pH value of soil, presence of water etc, in the crop field are tested using modem technology and appropriate crops are selected for cultivation.

Question 21.
What are the advantage of house farming and family farming
Answer:
a. Nontoxic food
b. Maximum utilization of land/space
c. Exercise
d. Mental Pleasure
e. Collaborative work of family members

HSSLive.Guru

Question 22.
How do microbial fertilizers help in plant growth?
Answer:
Microbial fertilizers contains micro organisms that increase the fertility of soil. They increase the amount of growth promoting factors in soil.
Eg: Rhizobium Azetobacter raise the amount of Nitrogen in the soil.

Question 23.
Which are the foreign varieties of fowl reared in our place?
Answer:
White leghorn, Rhode Island Red, Plymouth Rock, New Hampshire

Question 24.
What are the diseases of fowls
Answer:
Ranikatt, Fowl Cholera, Salmonellosis, Diarrhoea, Chores (Bacteria) Aspergillusis (Fungus).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Students can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Plus One Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers

Question 1.
Which of the following is not a Dobereiner triad?
a) Cl, Br, I
b) Ca, Sr, Ba
c) Li, Na, K
d) Fe, Co, Ni
Answer:
d) Fe, Co, Ni

Question 2.
The elements of s-block and p-block are collectively called ___________
Answer:
Representative elements

Question 3.
The cause of periodicity of properties is
a) Increasing atomic radius
b) Increasing atomic weights
c) Number of electrons in the valence shell
d) The recurrence of similar outer electronic configuration
Answer:
d) The recurrence of similar outer electronic configuration

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 4.
Halogen with highest ionization enthalpy is ___________ .
Answer:
Fluorine

Question 5.
Which of the following represents the most electropositive element?
a) [He]2s1
b) [He]2s2
c) [Xe]6s1
d) [Xe]6s2
Answer:
c) [Xe]6s1

Question 6.
Second electron gain enthalpy is
Answer:
always positive

Question 7.
Correct order of polarising power is
a) Cs+ < K+ < Mg2+ < Al3+
b) Al3+ < Mg2 + K+ < Cs+
c) Mg2+ < Al3+ < K+ < Cs+
d) K+ < Cs+ < Mg2+ < Al3+
Answer:
a) Cs+ < K+ < Mg2+ < Al3+

Question 8.
The IUPAC name of the element with atomic number is 109 is ___________
Answer:
Une

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 9.
The size of iso electronic species F~, Ne and Na+ is affected by
a) Nuclear charge
b) Principal quantum number.
c) Electron – electron interaction in outer orbitals.
d) None of the factors because their size is the same.
Answer:
Nuclear charge as nuclear charge is high the size is small.

Question 10.
In transition elements the differentiating electron occupies (n-1)d sublevel in preference to ______________
Answer:
np level

Plus One Chemistry Classification of Elements and Periodicity in Properties Two Mark Questions and Answers

Question 1.
The arguments made by two students are as given:
Student 1: ‘Hydrogen belongs to Group 1.’
Student 2: ‘Hydrogen belongs to Group 17.’
1. Who is right?
2. What is your opinion?
Answer:
1. Nobody is right.

2. Hydrogen has a one s-electron and hence can be placed in group 1 (alkali metals). It can also • gain an electron to achieve a noble gas arrangement and hence it can behave similar to a group 17 (halogen family) element. Because it a special case, hydrogen is placed separately at the top of the Periodic Table.

Question 2.
Match the following:

Sodium f-block
Oxygen s-block
Uranium d-block
Silver p-block

Answer:
Sodium – s-block
Oxygen – p-block
Uranium – f-block
Silver – d-block

Question 3.

  1. Which one has greater size, Na or K?
  2. Justify your answer.

Answer:
1. K

2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

Question 4.
The general characteristics of a particular block of elements is as given:
They are highly electropositive, soft metals. They are good reducing agents. They lose the outermost electron(s) readily to form 1+ ion of 2+ ion.

  1. Which block has this general characteristics?
  2. Write down two general characteristics of p-block.

Answer:
1. s-block

2. The p-block contains metals, non-metals and metalloids. They form ionic as well as covalent compounds.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 5.
The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. Out of oxygen and sulphur which has greater negative value for electron gain enthalpy? Justify.
Answer:
Sulphur has greater negative value (-200 kJ mol1) for electron gain enthalpy compared to that of oxygen (-141 kJ mol1). This is because, due to smaller size of oxygen the added electron experiences much repulsion from the electrons present in the shell. Due to large size, the electron-electron repulsion is much less in sulphur.

Question 6.
Some elements are given. Li, Cs, Be, C, N, O F, I, Ne, Xe.

  1. Arrange the above elements in the increasing order of ionization enthalpy.
  2. Arrange the given elements in the decreasing order of negative electron gain enthalpy.

Answer:

  1. Cs < I < Li < Xe < Be < C < N < O < F < Ne
  2. Cl > F > O > N > C > Be > Li > I >C s > Xe > Ne

Plus One Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers

Question 1.
Analyse the given figure and answer the questions that follow.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 1

  1. What is meant by atomic radius?
  2. Explain covalent radius.
  3. Write down another two types of terms expressed as atomic size.

Answer:

  1. Atomic radius is defined as the distance from the centre of the nucleus of the atom to the outermost shell of electrons.
  2. Covalent radius is defined as one half of the distance between the centre of nuclei of two similar atoms bonded by a single covalent bond.
  3. Vander Waals’ radius, Metallic radius

Question 2.
Consider the statement: The element with 1s2
configuration belongs to the p-block.’

  1. Identify the element.
  2. Do you agree with this statement?
  3. Justify.

Answer:

  1. Helium
  2. Yes
  3. Strictly, helium belongs to the s-block but its positioning in the p-block along with other group 18 elements is justified because it has a completely filled valence shell (1s²) and as a result, exhibits characteristic of other noble gases.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 3.
The properties of elements are a periodic function of
their atomic weights.

  1. Who proposed this law?
  2. Can you see anything wrong in this law? If yes, justify your answer.
  3. State modem periodic law.

Answer:

  1. Mendeleev
  2. Yes, atomic number is the more fundamental property of an element than atomic mass.
  3. The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
1. Define ionisation enthalpy.
2. IE1 <IE2 <IE3
What is meant by this? Justify.
Answer:
1. Ionisation enthalpy is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

2. It shows the increasing order the magnitude of successive ionization enthalpies. As the positive charge of the ion increases, it becomes more difficult to remove the valence electron due to increase in effective nuclear charge.

Question 5.
Say whether the following are true or false:

  1. On moving across a period ionization enthalpy decreases.
  2. Mg is biggerthan Cl.
  3. Ionization enthalpy of Li is less than that of K.

Answer:

  1. False
  2. True
  3. False

Question 6.
Analyze the graph given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 2
1. Identify the graph.
2. Account forthe following observations:
i) ‘Ne’ has the maximum value of ∆iH.
ii) In the graph from Be to B, ∆iH decrease.
Answer:
1. Graph showing the variation of ionisation enthalpy (∆iH) with atomic number (Z) of the elements of second period.

2. i) ‘Ne’ is an inert gas and it has closed electron shell with stable octet electronic configuration. Hence it has the maximum ionisation enthalpy in the second period.
ii) Be has completely filled electronic configuration and a more stable. So ionization enthalpy is high

Question 7.
Study the graph and answer the questions that follow:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 3
1. On moving down a group what happens to electron gain enthalpy?
2. Why chlorine shows more negative electron gain enthalpy than fluorine? ,

Answer:
1. On moving down a group,electron gain enthalpy becomes less negative because the size of the atom increases and the added electron would be farther from the nucleus.

2. Due to small size of F, the added electron goes to the smaller 2p quantum level and suffers significant replusion from the other electrons present in this level. But due to big size of Cl, the added electron goes to the n = 3p quantum level and occupies a larger region of space and the electron-electron repulsion is much less.

Question 8.
Electron gain enthalpy is an important periodic property.
1. What is meant by electron gain enthalpy?
2. What are the factors affecting electron gain enthalpy?
3. How electron gain enthalpy varies on moving across a period? Justify.

Answer:
1. Electron gain enthalpy(∆egH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion.

2. Effective nuclear charge, atomic size, electronic configuration

3. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently, it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the nucleus. Thus more energy is released.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 9.
1. The electron gain enthalpies of Be and Mg are positive. What is your opinion? Justify.
2. Electron gain enthalpies of nobles gases have large positive values. Why?
Answer:
1. The statement is correct. Electron gain enthalpies of Be (+240 kJ/mol) and Mg (+230 kJ/mol) are positive because they have stable electronic configurations with fully filled s-orbitals (Be – 2s², Mg – 3s²). Hence, the gain of electron is highly endothermic.

2. Noble gases have stable octet electronic configuration of ns2 np6 (except He -1 s²) and they have practically no tendency to accept additional electron. They have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. Hence, energy has to be supplied to add an extra electron.

Question 10.
During a group discussion, a student argued that ionization enthalpy depends only upon electronic configuration.
1. Do you agree? Comment.
2. Define shielding effect/screening effect.
3. Is there any relation between ionization enthalpy and shielding effect/screening effect? Explain.
Answer:
1. No. In addition to electornic configuration, ionization enthalpy depends upon other factors like atomic size, nuclear charge and shielding effect/screening effect.

2. In multi-electron atoms, the nuclear charge experienced by the valence electron will be less than the actual charge on the nucleus because it is shielded by inner core of electrons. This is called shielding effect or screening effect.

3. Yes. Ionization enthalpy decreases with increase in shielding effect/screening effect of inner electrons. This is because as a result of shielding of the valence electron by the intervening core of electrons it experiences a net positive charge which is less than the actual charge of the nucleus. In general, shielding is effective when the orbitals in the inner shells are completely filled.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 11.
a) Which of the following has higher first ionization enthalpy, N or O? Justify.
b) Which one is bigger, For F ? Why?
Answer:
1. N. This is because in N, three 2p-electrons reside in different atomic orbitals in accordance with Hund’s rule (\(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)) whereas in O, two the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion (\(2p_{ x }^{ 2 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)). Consequently, it is easier to remove the fourth 2p-electron from O than it is, to remove one of the three 2p-electrons from N.

2. F (136 pm) is bigger than F (72 pm). An anion is bigger than its parent atom. Addition of one electron in F results in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, attraction between nucleus and the electrons decreases and hence size increases.

Question 12.
Electronegativity differs from electron gain enthalpy.

  1. Do you agree?
  2. What do you mean by electronegativity?

Answer:

  1. Yes.
  2. Electronegativity is defined as the tendency of an atom in a chemical compound to attract the shared pair of electrons to itself.

Question 13.
Ionization enthalpy is an important periodic property.
1. What is the unit in which it is expressed?
2. What are the factors influencing ionization enthalpy?
3. How ionisation enthalpy varies in the periodic table?
Answer:
1. kJ mol-1.

2. Atomic/ionic radius, nuclear charge, shielding effect/screening effect, penetration effect and electronic configuration.

3. Ionization enthalpy generally increases with increase in atomic number across a period due to regular increase in nuclear charge and decrease in atomic size. Thus, the attractive force between the nucleus and the electron cloud increases. Consequently, the electrons are more and more tightly bound to the nucleus.

Ionisation enthalpy generally decreases from top to bottom a group: This is because the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nucelar hold on the valence electron decreases gradually and ionization enthalpy decreases.

Question 14.
The second period elements show anomalous ‘ behaviour.
1. Give reason.
2. What are the anomalous properties of second period elements?
Answer:
1. The first element is each group belong to the second period. The difference in behaviour of these elements from the other elements of the same group can be attributed to the following factors:

  • Small atomic size
  • Large charge/radius ratio
  • High electronegativity
  • Absence of d-orbtials in the valence shell

2. The important anomalous properties of second period elements are: diagonal relationship, maximum covalence of four and ability to form pπ —pπ multiple bonds.

Question 15.
Atomic size, valency, ionization enthalpy, electron gain enthalpy and electronegativity are the important periodic properties of elements.
1. What do you mean by periodicity?
2. Periodic properties are directly related to ___________
Answer:
1. The periodical repetition of elements with similar properties after certain regular intervals when the elements are arranged in the order of increasing atomic numbers is called periodicity,

2. Electronic configuration.

Question 16.
1. Which is the element among alkali metals having lowest ionization enthalpy?
2. What is meant by valence of an element? How it varies in the periodic table?
3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 4

Identify the elements A, B, C and D, if the graph represents halogens.
Answer:
1. Fr

2. Valence of an element is the combining capacity of that element. In the case of representative elements the number of valence electrons increases from 1 to 8 on moving across a period, the valence to the element with respect to H and Cl increases from 1 to 4 and then decreases from 4 to zero. On moving down a group, the number of valence electrons remains same and, therefore, all the elements in a group exhibit same valence.

3. A = F, B = Cl, C = Br, D = I

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 17.
Li and Mg belonging to first and second group in periodic table respectively resemble each other in many respects.
1. Name the relationship.
2. B can only form [BF4] ion while Al can form [AIF6]3-, though both B and Al belong to group 13. Justify.
Answer:
1. Diagonal relationship.

2. This is because B, being a second period element has a maximum covalence of 4. It cannot expand its covalence beyond 4 due to absence of d-orbitals. But Al, being a third period element has vacant d-orbitals in its valence shell and hence can expand its covalence beyond 4.

Question 18.
During a group discussion a student argues that both oxidation state and valence are the same.
1. Do you agree?
2. Justify taking the case of [AICI(H2O)5]2+.
Answer:
1. No. Valence refers to the combining capacity of an element whereas oxidation state is the charge assigned to an element in a compound based on the assumption that the shared electron in a covalent bond belongs entirly to the more electronegative element.

2. In [AICI(H2O)5]2+the valence of Al is 6 while its oxidation state is +3.

Question 19.
Among the elements of the third period, identify the element

  1. With highest first ionization enthalpy.
  2. That is the most reactive metal.
  3. With the largest atomic radius.

Answer:

  1. Ar
  2. Na
  3. Na

Question 20.
A cation is smaller than the corresponding neutral atom while an anion is larger. Justify.
Answer:
A cation is smaller than its parent atom because it has fewer number of electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased . repulsion among the electrons and a decrease in effective nuclear charge.

Question 21.
1. How the metallic character varies in the periodic table?
2. Categorize the following oxides into acidic, basic, neutral and amphoteric:
Al2O3, Na2O, CO2, Cl2O7, MgO, CO, As2O3, N2O
Answer:
1. The metallic character decreases from left to right across the period due to increase in ionization enthalpy along a period which makes loss of electrons difficult. From top to bottom a group metallic character increases due to decrease in ionization enthalpy. Thus, metallic character decreases diagonally from left bottom to right top of the periodic table.

2. Acidic oxides: CO2, Cl2O7
Basic oxides: Na2O, MgO
Neutral oxides: CO, N2O
Amphoteric oxides: Al2O3, As2O3

Question 22.
A group of ions are given below:
Na+, Al3+, O2-, Ca2+, Mg2+, F, N3-, Br
1. Find the pair which is not isoelectronic.
2. Arrange the above ions in the increasing order of size.
Answer:
1. Ca2+ and Br
2. Al3+ < Mg2+ < Na+ < F < O2- < N3- < Ca2+ < Br

Plus One Chemistry Classification of Elements and Periodicity in Properties Four Mark Questions and Answers

Question 1.
Statement 1: ‘Atomic mass is the fundamental property of an element.’
Statement 2: ‘Atomic number is a more fundamental property of an element than its atomic mass.’
1. Which statement is correct? Justify your answer.
2. Name the scientist who proposed this statement? What observation led him to this conclusion?
Answer:
1. Statement 2. Atomic number indicates the number of electrons present in an element. Most of the chemical properties of an element depend on its electronic configuration,

2. Henry Moseley. He observed regularities in the characteristic X-ray spectra of the elements. A plot of √υ (where u is the frequency of the X-rays emitted) against atomic number (Z) gave a straight line and not the plot of √υ vs atomic mass.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 2.
Atoms possessing stable configuration have less tendency to loss electrons and consequently will have high value of ionization enthalpy.
1. Justify this statement by taking the case of half-filled and completely filled electronic configurations.
2. The noble gases have highest ionization enthalpies in each respective periods. Why?
Answer:
1. Atoms with half-filled and completely filled electronic configurations have extra stability due to symmetric distribution of electrons and maximum exchange energy. Hence, more energy is required for the removal of their electrons. Elements like N (1s² 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)), P (1s² 2s² 2p6 3s² \(3p_{ x }^{ 1 }3{ p }_{ y }^{ 1 }3{ p }_{ z }^{ 1 }\)) etc. possessing half-filled shells have high ionization enthalpies.
Elements like Be ((1s² 2s²), Mg(1s² 2s² 2p6 3s²) etc. having completely filled shells show high values of ionization enthalpy.

2. Noble gases have closed electron shells and very stable octet electronic configurations (except He). Hence, maximum amount of energy is required to remove their valence electron.

Question 3.
Mendeleev arranged the elements in the order of increasing atomic weights.
a) Write down the merits of Mendeleev’s periodic table.
b) What are the demerits of Mendeleev’s periodic table?
Answer:
a) Merits of Mendeleev’s periodic table:

  • Study of elements – Elements are classified into groups with similar properties, thus facilitating the study of properties of elements.
  • Prediction of new elements – Mendeleev left certain vacant places in his table which provided a clue for the discovery of new elements, e.g. Eka-AI (for Ga), Eka-Si (for Ge).
  • Determination of correct atomic weights – With the help of this table, doubtful atomic weights of some elements were corrected.

b) Demerits of Mendeleev’s periodic table:

  • Position of hydrogen was not certain.
  • Anomalous pairs of elements – Certain elements of higher atomic weight preceed those with lower atomic weight, e.g. I (at.wt. 127) was placed after Te (at.wt. 128).
  • Lanthanides and Actinides are not given proper places in this periodic table.
  • No proper position for isotopes.

Question 4.
1. Name any three numerical scales of electronegativity.
2. How electronegativity varies in the periodic table? Justify.
Answer:
1. Pauling scale, Mullimen-Jaffe scale, Allred- Rochow scale.
2. Electronegativity generally increases from left to right a period and decreases from top to bottom in a group. This is because, from left to right across a period atomic size decreases and attraction between the valence electrons and the nucleus increases. From top to bottom in a group atomic size increases and attraction between the valence electrons and the nucleus decreases.

Question 5.
1. First ionization enthalpy of Na is lower than that of Mg. But its second ionization enthalpy is higher than that of Mg. Explain.
2. Which one is smaller, Na or Na+? Give reason.
Answer:
1. By the removal of one electron from Na it gets
the stable octet configuration of Ne. But when the first electron is removed from Mg it gets the unstable configuration of Na. It requires more energy due to small size and greater nuclear charge of Mg. In the case of Na the second electron is to be removed from a stable octet configuration which requires more energy than the removal of second electron from Mg.

2. Na+ (95 pm) is smaller than Na (186 pm). Acation is smaller than its parent atom. Na+ has fewer number of electrons (10 electrons) compared to Na (11 electrons). But the nuclear charge remains the same in both. Thus, effective nuclear charge per electron is greater in Na+. Thus, the attraction between nucleus and the remaining electrons increases and size decreases.

Question 6.
Removal of electron becomes easier on moving down the group.
1. Comment the above statement based on ionization enthalpy.
2. How electronic configuration influences the ionization enthalpy value?
Answer:
1. On moving from top to bottom in a given group, size of the atom increases and ionisation enthalpy decreases. Hence, it becomes easier to remove the valence electron.

2. Atoms with octet configuration, half-filled and completely filled configurations have extra stability and hence have higher values of ionization enthalpy.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 7.
The energy released during the addition of an electron to an isolated neutral atom is called electron gain enthalpy.
1. Explain how electron gain enthalpy differ from electronegativity.
2. The second ionisation enthalpy of an element is always greater than the first ionisation enthalpy. Give reason.
Answer:
1. Electron gain enthalpy(AegH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion. It is a quantitative property of an isolated gaseous atom, which can be measured. Whereas, electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. It is not a measureable quantity.

2. This is because to remove second electron from a positively charged ion more amount of energy is required due to increase in effective nuclear charge.

Question 8.
The physical and chemical properties of elements are periodic functions of their atomic numbers.
1. The atomic number of an element ‘X’ is 19. Write the group number, period and block to which X’ belong in the periodic table.
2. Name the element with
i) highest electronegativity and
ii) highest electron gain enthalpy
Answer:
1. The element is K.
19K= 1s² 2s² 2p6 3s² 3p6 4s1
Group number = 1
Period number = 4
Block = s-block

2. i) Fluorine
ii) Chlorine

Plus One Chemistry Classification of Elements and Periodicity in Properties NCERT Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table? (2)
Answer:
The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of groups rather than individually.

Question 2.
What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? (2)
Answer:
Mendeleev Periodic Law states that the properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of arrangements of elements from atomic weight to atomic number.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 3.
Consider the following species. N3-, O2-, F, N2+, Mg2+ and Al3+ (2)
1. What is common in them?
2. Arrange them in order of increasing ionic radii.
Answer:
1. Each one of these ions contains 10 electrons and
hence these are isoelectronic ions,

2. The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N3-, O2-, F, Na+, Mg2+ and Al3+, nuclear charge increase in the order:
N3-< O2- < F < Na+ < Mg2+ < Al3+
Therefore, the ionic radii decrease in the order:
N3- > O2- > F > Na+ > Mg2+ > Al3+

Question 4.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons (1)
Answer:
Nuclear mass does not affect the valence shell. Thus, option (c) is the correct answer.

Question 5.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (1)
a) F > Cl > O > N
b) F > O > Cl > N
c) Cl > F > O > N
d)0>F>N>CI
Answer:
a) F > Cl > O > N
Across a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is F > Cl > O > N and the choice (a) is correct.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Students can Download Chapter 1 Some Basic Concepts of Chemistry Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Plus One Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Question 1.
Which of the following is a mixture?
a) Graphite
b) Sodium chloride
c) Distilled water
d) Steel
Answer:
d) Steel

Question 2.
1 µ g = __________ g
[10-3, 10-6, 10-9, 10-12] ‘
Answer:
10-6

Question 3.
The number of significant figures in 0.00503060 is __________ .
Answer:
6

Question 4.
The balancing of chemical equations is based on which of the following law?
a) Law of multiple proportions
b) Law of conservation of mass
c) Law of definite proportions
d) Gay-Lussac law
Answer:
b) Law of conservation of mass

Question 5.
Which among the following is the heaviest?
a) 1 mole of oxygen
b) 1 molecule of sulfur trioxide
c) 100 u of uranium
d) 44 g carbon dioxide
Answer:
d) 44 g carbon dioxide

Question 6.
Calculate the number of atoms in 48 g of He?
Answer:
Gram atomic mass of He = 4 g.
Thus, numberofatomsin4g (1 mol) He = 6.02 × 1023
So number of atoms in 48 g of He = \(\frac{48}{4}\) × 6.02 × 1023
=12 × 6.02 × 1023
= 7.224 × 1024

Question 7.
One mole of CO2 contains how many gram atoms?
Answer:
3 gram atoms.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 8.
The ratio of gram atoms of Au and Cu in 22ct gold is __________
Answer:
7 : 2

Question 9.
A compound contains 69.5% oxygen and 30.5% nitrogen and its molecular weight is 92. The compound will be
Answer:
N2O4

Question 10.
The total number of electrons present in 1 mole of water is
Answer:
6 × 1024.

Question 11.
40g NaOH is present in 100 ml of a solution. Its molarity is __________
Answer:
10 M

Plus One Chemistry Some Basic Concepts of Chemistry Two Mark Questions and Answers

Question 1.
Classify the following substances into homogeneous and heterogeneous mixtures.

  • Milk
  • Iron
  • Air
  • Gasoline
  • Kerosene
  • Muddy water

Answer:

Homogeneous Heterogeneous
Milk, Iron
Gasoline
Air
Kerosene
Muddy Water

Question 2.
Calculate the volume occupied by 4.4 g of CO2 at STP?
Answer:
1 mole CO2 = 44 g
4.4 CO2 = 0.1 mole CO2
Volume occupied by 1 mol CO2 at STP = 22.4 L
∴ Volume occupied 0.1 mol CO2 at STP = 0.1 × 22.4 L
= 2.24 L

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 3.
During a group discussion a student argued that “the water of sea and river should have different chemical composition”.

  1. What is your opinion?
  2. Which law would you suggest to support your answer?
  3. State the law.

Answer:

  1. I can’t join with him.
    The water of sea and water of river must have the same chemical composition.
  2. Law of definite proportions.
  3. A given compound always contains exactly the same proportion of elements by weight.

Question 4.
“When science developed some theories are also modified”.
Write the modified atomic theory.
Answer:

  1. Atom is no longer considered as indivisible, it has been found that atom is made up of sub atomic particles called protons, neutrons and electrons.
  2. Atoms of the same element may not be similar in all respects.
  3. Atoms of different elements may be similar in one or more respects.
  4. The ratio in which atomic unit may be fixed and integral but may not be simple.
  5. The mass of atom can be changed into energy.

Question 5.
Carbon combines with oxygen to form CO and CO2.

  1. What is the law behind this?
  2. State the law.

Answer:

  1. Law of multiple proportions.
  2. If two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Question 6.
Calculate the volume occupied by 6.02×1025 molecules of oxygen at STP.
Answer:
Volume occupied by 1 mole of oxygen gas at STP = 22.4 l
i.e., Volume occupied by 6.02×1023 molecules of oxygen gas at STP = 22.4 l
Hence the volume occupied by 6.02 × 1025 molecules
of oxygen gas at STP = \(\frac{22.4 \times 6.02 \times 10^{25}}{6.02 \times 10^{23}}\) = 2240 l.

Question 7.
Calculate the molality of a solution of NaOH containing 20g of NaOH in 400 g solvent.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 1

Question 8.
Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 2

Question 9.
12 g of carbon reacts with 32 g of oxygen to form 44g of carbon dioxide.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Law of conservation of mass.
  2. Matter can neither be created nor destroyed. Or, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

Question 10.
When hydrogen and oxygen combine to form water, the ratio between volume of reactants and products is 2:1:2.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Gay Lussac’s law of gaseous volumes.
  2. When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Question 11.
Carbon form two oxides, the first contains 42.9% C and the second contains 27.3% carbon. Show that these are in agreement with the law of multiple proportions.
Answer:
In the first compound:
C = 42.9%
O = 100-42.9 = 57.1%
So, the ratio between the masses of C and O = 42.9:57.1 = 1:1.33
In the second compound:
C = 27.3%
O= 100-27.3= 72.7%
So, the ratio between the masses of C and O = 27.3 : 72.7= 1:2.66
Hence, the ratio of masses of oxygen which combines with a fixed mass of carbon is 1.33:2.66 or 1:2, a simple whole number ratio. This illustrates the law of multiple proportions.

Question 12.
Match the following:

A B
1 amu 1.008 x 1.66 x1024
Mass of 1 H atom 6.02 x 1023
Molar volume of O2 at STP 11.2L
Volume of 14g of N2 at STP 1.66 x 1024
Avogadro number 22.4L

Answer:

A B
1 amu 1.66 x 1024
Mass of 1 H atom 1.008 x 1.6 x1024
Molar volume of O2 at STP 22.4L
Volume of 14g of N2 at STP 11.2L
Avogadro number 6.02 x 1023

Question 13.
Calculate the molality of a solution containing 10 g ofNaOH in 200 cm3 of solution. Density of solution is 1.4 g/mL. (Molar mass of NaOH = 40)
Answer:
Mass of the solution = 200 × 1.04 = 208 g
Mass of NaOH (WB) = 10g Molar mass of NaOH (MB) = 40 g mol-1
Mass of water (WA) = (208 -10) g = 198 g = 0.198 kg
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 3

Question 14.
Calculate the mass percentage of oxygen in CaCO3.
Answer:
Molecular mass of CaCO3 = 100 g mol-1
Mass of oxygen in 100 g CaCO3=3 × 16 g = 48 g
Percentage of oxygen in CaCO3 =\(\frac{48}{100}\)×100 = 48%

Question 15.
KCIO3 on heating decomposes to KCI and O2. Calculate the mass and volume of O2 produced by heating 50 g of KCIO3.
Answer:
The reaction is represented as,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 4
According to the equation, 96 g of oxygen is obtained from 245 g of KCIO3.
Hence mass of oxygen obtained from 50 g KCIO3 is \(\frac{96 \times 50}{245}=19.6 \mathrm{g}\)
According to the equation 245 g of KCIO3 gives 3 moles of O2 at STP which is 3 × 22.4 L = 67.2 L
Volume of oxygen liberated by 50g of KCIO3
= \(\frac{67.2 \times 50}{245}=13.71 \mathrm{L}\)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 16.
Calculate the number of molecules present in

  1. 11g of CO2.
  2. 56 mL Of CO2 at STP.

Answer:
1. \(\frac{11}{44}\) = 0.25 mole
1 mole of CO2 contains 6.022 × 1023 molecules.
∴ 0.25 mole CO2 contains 6.022 × 1023 × 0.25
= 1.51 × 1023 molecules.

2. 56 mL = 0.056 L
\(\frac{0.056}{22.4}\) =0.0025 mole
= 6.022 × 1023 × 0.0025=1.5 × 1021 molecules

Question 17.
Calculate the number of moles of 02 required to produce 240 g of MgO by burning Mg metal. [Atomic mass: Mg=24, 0=16]
Answer:
2 Mg + O2 → 2MgO
No. of moles of MgO = \(\frac{240}{40}\)=6
No. of moles of 02 required = 6/2 = 3

Question 18.
Arrange the following in the increasing order of their mass.
(a) 1 g of Ca
(b) 12 amu of C
(c) 6.022 × 1023 mol-ecules of CO2
(d) 11.2 L of N2 at STP
Answer:
a) Mass of 1 g Ca = 1 g
b) Mass of 12 amu C = \(\frac{12}{6.022 \times 10^{23}}\) = 2 × 10-23
c) Mass of 6.022 × 1023 molecules of CO2 = 44 g
d) Mass of 11.2 L of N2 at NTP = \(\frac{28 \times 11.2}{22.4}\) = 14 g
(b) < (a) < (d) < (c)

Question 19.
Complete the table:

42g N2 1.5 mole N2 33600mLN2 (STP)
16g 0:  – – – – mole O2 11.2 L of O2 (STP)
….g CO2 1 mole CO2 – – – – L of C O2 (STP)
28g CO 1 mole CO – – – – mLCO (STP)

Answer:

42g N2 1.5 Mole N2 33600mLN2 (STP)
16g O2 0.5 Mole O2 11.2 L of O2 (STP)
44 g CO2 1 mole CO2 22.4 L of CO2 (STP)
28g CO 1 mole CO 22400mLCO (STP)

Question 20.
1. Irrespective of the source, pure sample of H20 always contains 88.89% by mass of oxygen and 11.11% by mass of hydrogen.
a) Which law is illustrated here?
b) State the law.
2. Complete the table by filling in the blanks:

48 g O2 1.5 mol O2 ……mL O2 (at STP)
…… g Na 2 gram atom Na 2NA Na atoms
…….g CO2 2.5 mol CO2 56 L (at STP)
8.5 g NH3 ……mol NH3 11.2 L (at STP)

Answer:
1. a) Law of definite proportions.
b) The same chemical compound always contains the same elements combined in the same fixed proprotion by mass.
2.

48 g O2 1.5 mol O2 33600 mL O2 (at STP)
46g Na 2 gram atom Na 2Na Na atoms
110 g CO2 2.5 mol CO2 56 L (at STP)
8.5 g NH3 0.5 mol NH3 11.2L(atSTP)

Question 21.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?.
Answer:
Mass of HCI in 0.25 mL 0.75 M HCl \(=\frac{36.5 \times 0.75 \times 0.25}{1000}=0.6844 \mathrm{g}\)
As per reaction 100 g CaCO3 reacts with 2 × 36.5 = 73 g of HCl
∴ Mass of CaCO3 reacting with 0.6844 g HCl \(=\frac{100 \times 0.6844}{73}=0.9375 \mathrm{g}\)

Plus One Chemistry Some Basic Concepts of Chemistry Three Mark Questions and Answers

Question 1.
During a Seminar, a student remarked that “Dalton’s atomic theory has some faulty assumptions”.
a) Do you agree with him?
b) What is the present status of Dalton’s atomic theory?
c) Write any two wrong postulates of Dalton’s atomic theory.
Answer:
a) I agree with him. Out of 6 Dalton’s postulates, 5 postulates are faulty and only one is correct.
b) Dalton’s atomic theory has undergone many modifications.
c)

  • All substances are made up of small indivisible particles called atoms.
  • Atoms of the same elements are identical in mass and other properties.

Question 2.
One gram atom of an element contains 6.023 × 1023 atoms.

  1. Find the number of atoms in 8 g oxygen.
  2. Which is heavier, 1 oxygen atom or 10 hydrogen atoms?
  3. Define mole and Avogadro number.

Answer:
1. 16 g oxygen contains 6.022 × 1023 atoms
∴ 8 g oxygen contains \(\frac{6.022 \times 10^{23}}{2}\) = 3.011 × 1023

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 5

3. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
Avogadro number – It is the number of discrete particles present in 1 mole of any substsnce. (Avogadro number, NA = 6.022 × 1023)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 3.

  1. Classify the following as homogeneous and heterogeneous mixtures.
    Air, Smoke, Gunpowder, NaCI solution, Petrol, Bronze, Mixture of sugar and sand.
  2. State and explain law of multiple proportions with example.

Answer:
1. Homogeneous-Air, NaCI solution, Bronze, Gun powder, Petrol.
Heterogeneous – Mixture of sugar and sand, Smoke.

2. When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other element bear a simple ratio. ,
eg. Carbon reacts with oxygen to form two compounds viz. CO and CO2. In CO mass ratio is 12:16. In CO2 mass ratio is 12:32. Then mass ratio between oxygen in the 2 compounds is 16:32 or 1:2 which is a simple whole number ratio. Hence, the law is verified.

Question 4.
1. One mole of an ideal gas occupies 22.4 L at STP
a) Calculate the mass of 11.2 L of oxygen gas at STP.
b) Calculate the number of atoms present in the above sample.
2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation.
N2 + 3H2 → 2NH3
Calculate the maximum amount of ammonia that can be formed.
Answer:
1. a) Mass of 22.4 L oxygen at STP = 32 g
∴ Mass of 11.2 L oxygen at STP = 16 g
b) No. of atoms present in 16 g of O2
\(\frac{6.02 \times 10^{23}}{2}\) ×2 = 6.02 × 1023 atoms

2. N2 + 3H2 → 2NH3
1 mole N2 + 3 mole H2 → 2moles NH3
1 mole N2 requires 3 mol H2
i.e., 28g N2 requires 6 g H2
Hence, 21 g N2 requires \(\frac{6 \times 21}{28}\) = 4.5 g H2
21 g N2 reacts completely and 0.5g H2 remains unreacted.
Hence, N2 is the limiting reagent.
28g N2 gives 2 × 17 g NH3
∴ 21 g N2 gives \(\frac{2 \times 17 \times 21}{28}\) = 25.5 g NH3

Question 5.
When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other bear a simple ratio.
i) Name the above law.
ii) Explain the above law by taking oxides of carbon.
Answer:
i) Law of multiple proportions.
ii) Carbon reacts with oxygen to form two compounds viz. CO and CO2.
In CO, mass ratio is 12:16
In CO2,mass ratio is 12:32
Ratio of the masses of oxygen combining with a fixed mass of carbon in the two compounds is 16:32 or 1:2, which is a simple whole number ratio.

Question 6.
A compound contains 80% carbon and 20% hydrogen. If the molecular mass is 30 calculate empirical formula and molecular formula.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 6

Question 7.
A compound contains 4.07% of hydrogen, 24.27% of carbon and 71.65% of chlorine. The molar mass is 98.96. What is the empirical and molecular formula?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 7

Question 8.
Nitrogen forms various oxides.
1. Identify the law of chemical combination illustrated here. Also state the law.
2. Determine the formula of each oxide from the given data and illustrate the law.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 8

Answer:
1. Law of multiple proportions.
When two elements combine to form more than one compound the different mass of one of the elements which combine with the fixed mass of the other element bear a simple ratio.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 9
In NO and NO2, the masses of oxygen combining with a fixed mass (14 g) of nitrogen are in the ratio, 16:32 = 1:2. Similarly, in N2O and N2O3, the masses of oxygen combining with a fixed mass (28 g) of nitrogen are in the ratio, 16:48 = 1:3. These are simple whole number ratios. Hence, the law of multiple poportions is verified.

Plus One Chemistry Some Basic Concepts of Chemistry Four Mark Questions and Answers

Question 1.
Which of the following weighs more?
a) 1 mole of glucose
b) 4 moles of oxygen
c) 6 moles of N
d) 5 moles of sodium
Answer:
a) 1 mole glucose = (72 + 12 + 96) g = 180 g
b) 4 moles of oxygen = 4 × 32g = 128g
c) 6 moles of nitrogen = 6 × 14 g = 84 g
d) 5 moles of Na = 5 × 23 g = 115 g.
Thus, 1 mole glucose weighs more.

Question 2.
3 g of H2 is mixed with 29 g of O2 to yield water.
1. Which is the limiting reagent?
2. Calculate the maximum amount of water that can be formed.
3. Calculate the amount of the reactants which remains unreacted.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 10
According to the equation, 4 g H2 requires 32 g
So 3 g H2 requires \(\frac{377 \times 33}{44}\) = 24 g O2.
Here 3 g H2 is mixed with 29 g of O2. All H2 will react. Hence H2 is the limiting reagent.

2. According to the equation, 4 g H2 gives 36 g H2O. Hence 3 g H2 will give 36 × 3/4 = 27 g H2O.

3. Amount of O2 unreacted = (29 – 24)g = 5 g

Question 3.
a) Calculate the mass of oxygen required for the complete burning of 2 g of carbon.
b) Calculate the molar mass of (i) CO2 (ii) CH4
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 11

Question 4.
One gram mole of a substance contains 6.022 x 1023 molecules.
1. 24 g of carbon is treated with 72 g of oxygen to form CO2. Identify the limiting reagent.
2. Find the number of molecules of CO2 formed in this situation.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 12

2 mol of C requires 2 mol of O2.
2 mol C completely reacts with 2 mol of O2 and 0.25 mol O2 and 0.25 mol O2 remains unreacted. Hence, C is the limiting reagent.

2. No. of moles of CO2 formed = 2
∴ No. of molecules of CO2 formed
= 2 × 6.022 × 1023 = 1.2044 × 1024

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 5.
One gram mole of a substance contains 6.022×1023 molecules.
i) Find out the number of molecules in 2.8 g of nitrogen.
ii) Which is the heavier-one SO2 molecule or one CO2 molecule?
Answer:
i) No. of molecules in1 mole of N2 = 6.022 × 1023
i.e., No. of molecules in 28 g of N2 = 6.022 × 1023
∴ No. of molecules in 2.8 g N2
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 13

Question 6.
a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal?
b) Match the following:
1/12th the mass of C12 atom – 1 mole
1 g of hydrogen atom – amu
22.4 L O2 at NTP – gram mole
180 g of glucose – gram atom
6.022 × 1023 particles – molar volume
Answer:
a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively.
First Oxide :
Mass of oxygen = 100 – 78.7 = 21.3 g
No. of parts by mass of oxygen combining with one part by mass of metal =\(\frac{78.7}{21.3}=3.7 \mathrm{g}\)

Second oxide:
Mass of oxygen = 100 – 64.5 = 35.5 g
No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}=1.9 \mathrm{g}\)
The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

b) 1/12th the mass C12 atom – amu
1 g of hydrogen atom – gram atom
22.4 L O2 at NTP – molar volume
180 g of glucose – gram mole
6.022 × 1023 particles – 1 mole

Question 7.
Calculate
1. The number of molecules present in 1 g of water.
2. The volume of 0.2 mole of sulphur dioxide at STP.
Answer:
1. Number of moles in 1 g water = \(\frac{1}{8}\)
∴ No. of molecules in 1 g water
\(=\frac{1 \times 6.022 \times 10^{23}}{18}=3.35 \times 10^{22}\)

2. Volume of 0.2 mol S02 at STP = 0.2 × 22.4 litre
= 4.48 litre

Question 8.
“One mole of all substances contain the same number of specified particles.”
a) Justify the statement.
b) Howto connect mole, gram mole, and gram atom?
c) What is the relation between number of moles and volume?
d) Calculate the number of moles of a gas in 11.2 L at • STP.
Answer:
a) This statement is true i.e., one mole of all sub-stances contain the same number of specified particles. According to Avogadro’s law.
1 mole of any substance contains 6.022 × 1023 specified particles.

b)
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 14
1 gram mole is the molecular mass expressed in gram. It is the mass of 1 mole molecules in gram. Thus, 1 gram mole contains 1 mole molecules.
1 gram atom is the atomic mass expressed in gram. It is the mass of 1 mole atoms in gram. Thus, 1 gram atom contains 1 mole atoms.

c) Number of moles is directly proportional to volume (according to Avogadro law).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 15

Plus One Chemistry Some Basic Concepts of Chemistry NCERT Questions and Answers

Question 1.
Calculate the molecular mass of the following : (3)
1. H2O
2. CO2
3. CH4
Answer:
1. Molecular mass of H2O = 2(1.008 u) +16.00 u
= 18.016u

2. Molecular mass of CO2 = 12.01 u + 2(16.00 u)
= 44.01 u

3. Molecular mass of CH4 = 12.01 u + 4(1.008 u)
= 16.042 u

Question 2.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 16

Question 3
Calculate the amount of carbon dioxide that could be produced when (3)
1. 1 mole of carbon is burnt in air.
2. 1 mole of carbon is burnt in 16 g of dioxygen.
3. 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
1. The balanced equation for the combustion of carbon dioxide in dioxygen in air is
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 17

In air combustion is complete.
Hence, Amount of CO2 produced when 1 mole of carbon is burnt in air = 44 g

2. As only 16 g dioxygen is available it is the limiting reagent.
Hence, amount of CO2 produced = \(\frac{44}{32}\)×16 = 22 g

3. Here again, dioxygen is the limiting reactant. Therefore, amount of CO2 produced from 16 g dioxygen = \(\frac{44}{32}\)×16 = 22g

Question 4.
Chlorine is prepared in the laboratory by treating manganase dioxide (MnO2) with aqueous hydrochloric acid according to the reaction,
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
(Atomic mass of Mn = 54.94 u) (2)
Answer:
1 mol of MnO2, i.e., 54.94 + 32 = 86.94 g MnO2 react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl.
∴ Mass of HCl reacting with 5.0 g of MnO2
\(=\frac{146}{86.94} \times 5 \mathrm{g}=8.4 \mathrm{g}\)

Question 5.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 18

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Students can Download Chapter 5 States of Matter Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Plus One Chemistry States of Matter One Mark Questions and Answers

Question 1.
When a gas is compressed at constant temperature
a) The speeds of the molecules increase
b) The collisions between the molecules increase
c) The speed of the molecules decrease
d) The collisions between the molecules decrease
Answer:
b) The collisions between the molecules increase

Question 2.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called ___________
Answer:
Boyle temperature or Boyle point

Question 3.
The compressibility factor is given by the expression
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 1
Answer:
a) \(\frac{p V}{n R T}\)

Question 4.
Two flasks of equal volume contain CO2 and SO2 respectively at 298 K and 1.5 atm pressure. Which of the following is equal in them?
a) Masses of the two gases
b) Rates of effusion
c) Number of molecules
d) Molecular structures
Answer:
c) Number of molecules

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 5.
With rise in temperature, viscosity of a liquid
a) Increases
b) Decreases
c) Remains constant
d) May increase or decrease
Answer:
b) Decreases

Question 6.
The unit of‘b’ in VanderWaals equation of state.
Answer:
l mol-1

Question 7.
Most probable velocity, average velocity, and root mean square velocity are related by
Answer:
1 : 1.128 : 1.224

Question 8.
The volume of 2.8g of CO at 27°C and 0.821 atm pressure is (R = 0.0821 l atm Km-1 ol-1)
Answer:
3L

Question 9.
The density of gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temp will its density become 0.75d?
Answer:
400K

Question 10.
The rms velocity of an ideal gas at 27°C is 0.3ms-1. ‘ Its rms velocity at 927°C in (ms-1) is
Answer:
0.6m/s

Plus One Chemistry States of Matter Two Mark Questions and Answers

Question 1.
Find out the relation between the first pair and complete the second pair.
a) Boyle’s law: Temperature
Charles’ law : ……………….
b) Avagadro’slaw: V α n
Ideal gas equation: ……………..
Answer:
a) Pressure
b) pV=nRT

Question 2.
The graphs of Boyle’s law as plotted by Student 1 (Graph 1) and Student 2 (Graph 2) are given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 2

  1. Which is the correct graph?
  2. Justify your answer.

Answer:

  1. Both the graphs are correct.
  2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 3.
The rate of diffusion of hydrogen is less than that of oxygen.

  1. Do you agree?
  2. Which law is applied here?
  3. State the law.

Answer:

  1. Yes.
  2. Graham’s law of diffusion.
  3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Question 4.
The ideal gas equation has been modified for real gases by applying pressure and volume corrections.

  1. What is the corrected equation known as?
  2. Write the equation and explain the terms.

Answer:

  1. The van derWaals’equation.
  2. \(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V — nb) = nRT
    where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Question 5.
‘Moist soil grains are pulled together.’

  1. Name the related phenomenon.
  2. Justify.

Answer:

  1. Surface tension.
  2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

Question 6.
1. What is aqueous tension?
2. What is its significance in the determination of pressure of a dry gas?
Answer:
1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature.

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total presssure of the moist gas.
Pdry gas=PTotal-Aqueous tensi0n

Question 7.
A balloon filled with air, when kept in sunlight bursts after some time.

  1. Name the related law.
  2. Justify.

Answer:

  1. Charles’ law
  2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the baloon expanded more than that the balloon could afford, it bursted.

Question 8.
Define surface energy. What is its SI unit?
Answer:
Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m-2.

Question 9.
a) Based on Boyle’s law how will you show that at a constant temperature, pressure is directly proportional to the density of a fixed mass of the gas?
b) Give the relation between density and molar mass of a gaseous substance.
Answer:
a) According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 3

Question 10.
The isotherm of carbon dioxide at various temperatures is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 4
1. What is the significance of the shaded area?
2. Identify the pressure at which liquid CO2 appears for the first time when cooled form 30.98 °C. What is this pressure called?
Answer:

  1. At any point in the dome shaped shaded area liquid and gaseous CO2 exists in equilibrium.
  2. 73 atm. Critical pressure (pc).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 11.
Certain properties of liquids are given below: Classify them on the basis of effect of temperature on them,

  1. Evaporation
  2. Vapour pressure
  3. Surface tension
  4. Viscosity

Answer:
Properties which increase with increase in temperature: Evaporation & Vapour pressure Properties which decrease with increase in temperature: Surface tension & Viscosity

Question 12.
The size of the water bubbles increases on moving to the surface.
1. Name the law responsible for this.
2. What is your justification?
Answer:
1. Boyle’s law.

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Question 13.
What are the properties of liquid state?
Answer:
Vapour pressure, Boiling point, Viscosity and Surface tension.

Plus One Chemistry States of Matter Three Mark Questions and Answers

Question 1
Analyse the following graph :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 5
1. Name the gas law associated with this graph.
2. State the law.
3. Give the mathematical expression of this law.
Answer:
1. Boyle’s law.

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversly with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n)
Or p = k × \(\frac{1}{V}\) where k is the proportionality
constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 2.
1. What are van der Waals’ forces?
2. Which are the different types of van der Waals’ forces?
3. Arrange the van der Waal’s forces in the increasing order of their strength.
Answer:
1. The attractive intermolecular forces are known as van der Waals’ forces.

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding

Question 3.
A graphical representation of Charles’ law is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 6

  1. What is the temperature corresponding to the point‘A’called? .
  2. What will be the temperature at that point A’ in degree Celsius?
  3. What is the significance of this temperature?

Answer:

  1. Absolute zero temperature
  2. -273.15 °C
  3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 4.
Assume that two gases X and Y at the same temperature and pressure have the same volume.
1. Which of the following is correct?
No. of moles of X= No.of moles of Y
No. of moles of X ≠ No.of moles of Y
2. Which law helped you to find the answer?
3. State the law.
Answer:
1. No. of moles of X= No.of moles of Y

2. Avogadro’s law

3. Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 5.
During a seminar session in the class, the presenter argued that equal amounts of both H2and N2 on heating at constant pressure will expand in the same rate. Another student objected this argument by saying that they will expand differently since their molecular masses are different.

  1. Who is correct in your opinion?
  2. Which law helped you to reach the answer?
  3. State the law and give its mathematical expression.

Answer:
1. The argument of the presenter is correct.

2. Charles’ law

3. Charles’ law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically, V ∝ T (at constant n and P) Or \(\frac{V}{T}\) = K, where K is the proportionality constant which depends on the pressure of the gas, its amount and the unit in which V is expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 6.
1. What is an ideal gas?
2. Give the ideal gas equation and explain the terms.
3. Derive the ideal gas equation.
Answer:
1. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly at all conditions is called an ideal gas.

2. The ideal gas equation is pV = nRT where p = pressure, V = volume, n = number of moles, R = universal gas constant and T = absolute temperature.

3. According to Boyle’s law:
V ∝ \(\frac{1}{p}\) at constant T and n.
According to Charles’ law:
V ∝ T. at constant n and p.
According to Avogadro law,
V ∝ n , at constant p and T Combining the above three equations,
V ∝ \(\frac{nT}{p}\)
⇒ V = R\(\frac{nT}{p}\), where R is the proportionality constant known as universal gas constant.
Or pV = nRT, the ideal gas equation.

Question 7.
Partial pressure of a vessel containing Cl2, CO2 and CO is the sum of the partial pressures of Cl2, O2 and CO.
1. If so, is it correct to say partial pressure of a vessel containing NH3 and HCl gases is the sum of their partial
pressures? Justify.
2. Which law helped you to answer this?
3. State the law.
Answer:
1. No. NH3 reacts with HCl to form NH4CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure.

2. Dalton’s law of partial pressures.

3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 8.
The average kinetic kenergy of the gas molecules is directly proportional to the absolute temperature.

  1. Which theory is related to this assumption?
  2. Write the other postulates of this theory.

Answer:
1. Kinetic moleculartheory of gases.

2.

  • The volume of a gas molecule is negligible when compared to the whole volume of the gas.
  • There is no force of attraction between the particles of a gas at ordinary temperature and pressure.
  • The gas molecules are in random motion.
  • During motion, they collide with each other and also with the walls of the container.
  • Gravity has no influence in the movement of gas molecules.
  • Pressure of a gas is due to the collision of gas molecules with the walls of the container.
  • At any particular time, different particles in the gas have different speeds and hence different kinetic energies.

Question 9.
Three mateorological baloons filled with equal amount of helium, rising in the atmosphere are shown below: (Assume that temperature remains constant).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 7

  1. Which of the baloons will be at the lowest altitude?
  2. Which law helped you to find the answer?
  3. State the law.

Answer:

  1. C –
  2. Boyle’s law.
  3. At low altitudes pressure is high. According to Boyle’s law for a given mass of a gas, greater the pressure lower is the volume at constant temperature.

Question 10.
‘All the postulates of the kinetic molecular theory of gases are correct.’

  1. Do you agree with the statement?
  2. If no, write the wrong postulates of this theory.
  3. Give justification.

Answer:
1. No.

2.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  •  There is no force of attraction between the molecules of a gas.

3. If assumption i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens.
If assumption ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

Question 11.
Two gases with equal molecular mass will have the same rate of diffusion.’

  1. Do you agree?
  2. Explain.
  3. Substantiate your answer with an example.

Answer:
1. Yes.

2. According to Graham’s law of diffusion the rate of diffusion depends only on the molecular mass. So if the molecular masses are the same, their rate of diffusion is same.

3. Both CO and N2 have the same molecular mass (28 g mol-1)
Rate of diffusion of CO = Rate of diffusion of N2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 12.
Water can be boiled more quickly on the top of a mountain.

  1. Do you agree?
  2. What is the reason?
  3. What is called boiling point of a liquid?
  4. How normal boiling point and standard boiling point differ?

Answer:
1. Yes.

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly.

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure.

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

Question 13.
Ethanol flows faster than honey.

  1. Name the related phenomenon.
  2. Explain this phenomenon.
  3. What is the effect of temperature on this?

Answer:
1. Viscosity.

2. Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.

3. Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

Question 14.
Liquid drops attain spherical shape.

  1. Which property of liquids is responsible for this?
  2. Explain the phenomenon and justify.
  3. Suggest another consequence of this phenomenon.

Answer:
1. Surface tension.

2. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. The lowest energy state of the liquid will be when surface area is minimum. Spherical shape satisfies this condition.

3. Fire polishing of glass – On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges due to surface tension, which makes the edges smooth.

Question 15.
Vapour pressure is an important property of liquids.

  1. What is vapour pressure?
  2. How boiling point and vapour pressure are related?
  3. Pressure cooker is used for cooking food at higher altitudes. Give reason.

Answer:
1. Vapour pressure of a liquid is the pressure exerted by the vapour which is in equilibrium with liquid at a given temperature.

2. Boiling point of a liquid is the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. Thus, lower the vapour pressure of a liquid higher will be its boiling point and vice-versa.

3. At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. In a pressure cooker, the internal pressure is greater than atmospheric pressure. Hence, in a pressure cooker water boils at a temperature higher than its normal boiling point of 100 °C. Thus, cooking becomes more effective.

Question 16.
Assume that ‘A’, ‘B’ and ‘C’ are three non-reacting gases kept in a vessel at a constant temperature.
Then, PTotal= PA + PB + PC
1. Name the related law.
2. How can you explain the above law on the basis of kinetic molecular theory of gases?
Answer:
1. Dalton’s law of partial pressures.

2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases.

Question 17.
1. Write the general equation which relates the different variables of a gas used to describe the state of any ideal gas.
2. A flask at 295 K contains a gaseous mixture of N2 and O2 at a total pressure of 1.8 atm. If 0.2 moels of N2 and 0.6 moles of O2 are present, find the partial pressures of N2 and O2.
3. What is meant by Boyle temperature or Boyle point?
Answer:
1. PV = nRT
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 8

3. It is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.

Question 18.
1. Liquid tries to rise or fall in the capillary. Name the related phenomenon.
2. What is the effect of temperature on the above phenomenon?
3. WhatistheSI unit of the above phenomenon.
Answer:
1. Surface tension.

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

3. N m-1.

Question 19.
1. Define critical temperature (Tc ).
2. CO2 cannot be liquified above 31.1°C. Why?
3. The critical temperatures of ammonia and carbon dioxide are 405.5 Kand 304.10 K respectively. On cooling, which of these gases will liquify first? Justify.
Answer:
1. It is the highest temperature at which a gas can
be liquified by applying external pressure.

2. The critical temperature (Tc ) of CO2 is 30.98°C. This is the highest temperature at which liquid CO2 is observed. Above this temperature it is gas.

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 20.
a) Will water boils at higher temperature at sea level or at top of a mountain. Explain.
b) A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer:
a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 9

Question 21.
Real gases deviate from ideal behaviour.
1. What are the two wrong postulates of kinetic theory of gases, responsible for deviation of real gases from ideal behaviour?.
2. When do real gases deviate from ideal behaviour?
Answer:
1.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  • There is no force of attraction between the molecules of a gas.

2. Real gases deviate from ideal behaviour at high pressure and low temperature, when the gas molecules are very close to each other.

Question 22.
1. What is meant by compressibility factor, Z?
2. What is the significance of compressibility factor?
3. A plot of pV/nRT of oxygen gas against p is as follows:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 10
a) Is the gas ideal or real?
b) Write the equation of state of the above gas.
Answer:
1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(z=\frac{p V}{n R T}\).

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor.

3. a) Real gas.
\(\left[p+\frac{a n^{2}}{V^{2}}\right]\)[V-nb] = nRT (van der Waals’ equation)

Question 23.
a) What is the difference between gas and vapour?
b) Analyse the vapour pressure vs temperature curve shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 11
Arrange the compounds shown in the graph in the decreasing order of their normal boiling points,

c) The density of a gas was found to be 2.92 g L1 at 27 °C and 2.0 atm. Calculate the molar mass of the gas.
Answer:
a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance.

b) water > ethyl alcohol > carbon tetrachloride > diethyl ether
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 12

Question 24.
1. How will you account for the observation that automobile tyre is inflated with lesser air in summer than in winter?
2. A sample of gas occupies 250 mLat27 °C. What volume will it occupy at 35 °C if there is no change in pressure?
Answer:
1. This can be explained on the basis of Gay Lussac’s law, according to which at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. In summer season the temperature will be higher. Hence, pressure will increase and the tyre may burst if filled with more air. But during winter temperature is low and hence pressure will below.

2. According to Charles’ law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 13

Question 25.
Real gases behave ideally at low temperature and high pressure.

  1. Is the above statement correct or not?
  2. Justify.
  3. Write the van der Waals’ equation for 1 mole of a real gas.

Answer:
1. The statement is wrong.

2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart.

3. \(\left(p+\frac{a}{V^{2}}\right)\)(V – b) = RT

Question 26.
1. Distinguish between real gas and ideal gas.
2. Explain the deviation of the following gases from ideal behaviouron the basis of the pV vs. p plot. CO, CH4, H2 and He.
Answer:
1. Real gas do not follow, Boyle’s law, Charles law, and Avagadro law perfectly under all conditions. Ideal gas follow, Boyle’s law, Charles’ Law and Avagadro law strictly under all conditions.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 14
It can be seen that at constant temperature pV vs p plots for these gases are not straight lines. Two types of curves are seen. In the curves for H2 and He, as the pressure increases the value of pV also increases. These gases show positive deviation from ideal behaviour at all pressures. The second type of plot is seen in the case of CO and CH4. For these gases the pV value decreases with increase in pressure and reaches to a minimum value characteristics of the gas. After that PV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously.

Question 27.
1. What is meant by laminar flow?
2. Derive the expression for the force responsible for flow of layers of a liquid.
Answer:
1. When a liquid flows overa fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

2. Consider three layers of a flowing liquid as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 15
For any layer, the layer above it accelerates its flow and the layer below this retards its flow. If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by \(\frac{du}{dz}\). A force is required to maintain the
flow of layers. This force is proportional to the area of contact (A) of layers and velocity gradient.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 16
where η is a proportionality constant called coefficient of viscosity.

Question 28.
1. What are London forces?
2. What is the relation between London forces and the distance between the particles?
Answer:
1. The attractive force between two temporary dipoles is known as London forces or Dispersion forces.

2. London forces are always attractive and the interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.

Question 29.
1. What is meant by thermal energy and thermal motion?
2. Can oxygen exist as a gas at -273.15°C? Write the significance of this temperature.
Answer:
1. Thermal energy is the energy of a body arising from motion of its atoms or molecules. The movement of particles due to thermal energy is called thermal motion.

2. At – 273.15 °C oxygen will not exist as a gas. In fact all the gases get liquified before this temperature is reached. It is the absolute zero of temperature, which is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 30.
Molecues of a gas are in a state of continuous motion
1. What is most probable speed?
2. Give the equation for average speed of molecules.
Answer:
1. Most probable speed is the speed possessed by the maximum fraction of molecules of the gas at a given temperature.

2. If there are ‘n’ molecules in a sample and their individual speeds are u1, u2 ……… un, then, average speed of molecules, uav is given by the equation:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots+u_{n}}{n}\)

Plus One Chemistry States of Matter Four Mark Questions and Answers

Question 1.
1. State the Avogadro law.
2. Give the mathematical expression of this law.
3. What is the value of molar volume of an ideal gas at 273.15 K and 1 bar?
4. Show that, at constant temperature and pressure, the density of an ideal gas is proportional to its molar volume.
Answer:
1. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

2. v ∝ n (at constant P and T)
⇒ V = k × n where k is a proportionality constant.

3. 22.71098 L

4. According to Avogadro law, for n moles of an ideal gas,
V = k × n
But n = \(\frac{m}{M}\) where m is the mass of the gas and M is its molar mass.
Thus, V = k × \(\frac{m}{M}\)
On rearranging the above equation,
M = k × \(\frac{m}{V}\) = k × d
⇒ M ∝ d

Question 2.
The speed of molecules is a measure of their average kinetic energy.
a) What is root mean square speed?
b) Give the equation for root mean square speed.
c) Calculate the following:
i) Root mean square speed of methane molecule at 27°C.
ii) Most probable speed of nitrogen molecule at 25°C
Answer:
a) It is the square root of the mean of the squares of speeds of various molecules of the gas at a given temperature.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 17

Question 3.
The graph A is drawn at high temperature and low pressure and graph B is drawn at low temperature and high pressure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 18
a) Which graph represents ideal behaviour?
b) Give the equation for combined gas law.
c) A baloon occupies volume of 700 mL at 25°C and 760 mm of pressure. What will be its volume at higher attitude when temperature is 15°C and pressure is 600 mm Hg.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 19

Question 4.
1. Give the relationship among the three types of molecular speeds.
2. Drawthe Maxwell-Boltzmann distribution showing all the molecularspeeds.
3. Which of the following molecules will have the higher value of most probable speed at the same temperature, N2 or Cl2? Justify.
Answer:
1. Root mean square speed, average speed and the most probable speed have the following relationship:
Urms > Uav > Ump
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 20
3. N2. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules.

Plus One Chemistry States of Matter NCERT Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C? (2)
Answer:
p1 = 1 bar p2 = ?
V1 = 500 dm³ V2 = 200 dm³
Temperature remains constant.
According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 21

Question 2.
Using the equation of state pV=nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. (2)
Answer:
According to ideal gas equation :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 22

Question 3.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP. (3)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 23

Question 4.
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure (R = 0.083 bar L K-1 mol-1). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 24

Question 5.
Explain the significance of van der Waal parameters. (2)
Answer:
The van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effectve size of the gas molecules.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion

You can Download Motion Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion

We can state whether an object is in a state of rest or motion only with reference to another object. The object which is taken as reference is the reference body. Reference body is the object with respect to which the state of rest or motion of an object is described.

Distance and Displacement

If an object travels from initial position to final position through different routes the distance will not be equal. But if it travels through straight line the distance is called displacement. The distance has only magnitude. But displacement has both magnitude and direction.

While stating the displacement, it is necessary to indicate the direction along with the magnitude of the distance travelled. Such physical quantities having both magnitude and direction are referred to as vector quantities. Physical quantities of which the direction is not to be indicated, are scalar quantities When a body travels along a straight line in the same direction, the magnitude of its distance and displacement will be equal.

Speed and Velocity

Speed is the distance travelled in unit time. Velocity is the displacement in unit time.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 1

The unit of speed and velocity is metre/second (m/s)

Uniform speed and Non uniform speed

If a body in motion covers equal distance in equal intervals of time the body is said to have uniform speed. That is if a car travels 20 km in first 2 seconds, again travel 20 km in successive 2 seconds, the body is said to have uniform speed. But the body travels 20 km in first 2 seconds, 40 km in next 2 seconds and then travels 50 km in next 2 seconds the body is said to have non uniform speed.

The speed of a car in non uniform speed is different. So to find out the speed calculate the average speed
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 2

Uniform velocity and Non uniform velocity

A body has uniform velocity if it covers equal displacements in the same direction in equal intervals of time. But if it is not in equal velocity in equal interval it is in non uniform velocity.

Acceleration

A body with a velocity of o m/s takes 5 seconds to rise the velocity to 5 m/s the change of velocity is 5 – o = 5 m/ s. The change is taken place in 5 sec. \(\frac{change of velocity}{time}\) the rate of change of velocity which is called acceleration.

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Retardation

 

When the velocity of a body is decreasing the acceleration is negative. This type of acceleration is called retardation.

Carelessness of pedestrians is also one of the reasons for road accidents.
To reduce the accidents following precuations are to be taken.

  • walk along the footpath only
  • walk only along the right side of the road
  • cross roads only at the zebra crossing.
  • while walking along the road late in the evening or night avoid wearing black or dark coloured clothes.

Motion Textbook Questions and Answers

Question 1.
Which of the following does not belong to the group?
(velocity, acceleration, speed, displacement)
Answer:
speed

Question 2.
The statement of a child is as follows: “My displacement is zero though I ran 250 m”. What is meant by this?
Answer:
The child reaches the point which he started. So the displacement is o.

Question 3.
All objects having uniform speed need not have uniform velocity. Describe with the help of examples.
Answer:
If a body is in uniform velocity the direction and quantity will be equal.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 3
In figure (1) from A to B and B to C the direction is different. Therefore it is not in uniform velocity. In figure (2) it is in uniform velocity.

Question 4.
Bus A covered 75 m in 5 s. Bus B covered 169 m in 13 s
a. Which bus covered a greater distance?
b. Which bus has a higher speed?
Answer:
The bus B covered more distance.
b. Speed = \(\frac{distance}{time}\)
Speed of A= \(\frac{75}{5}\) = 15 m/s
Speed of B = \(\frac{169}{13}\) = 13 m/s
So the bus A has higher speed

Question 5.
What is the acceleration of a car which started from rest and acquired a velocity of 40 m/s in 8 s?
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 4

Question 6.
A car covered the first 400 m distance with a speed of 8 m/s, the next 1200 111 with a speed 12 m/s and the last 360 m with a speed of 12 m/s. Calculate the average speed of the car.
Answer:
Average speed = \(\frac{Total distance travelled}{time taken to trvel the distance}\)
Time taken to travel the distance 400m = \(\frac{400}{8}\) = 50s
Time taken to travel the distance 1200m = \(\frac{1200}{10}\) = 120s
Time taken to travel the distance 360m = \(\frac{360}{12}\) = 30s
Total distance travelled = 400 + 1200 + 360 = 1960m
Total time to travel = 50 + 120 + 30 = 200s
Average speed=\(\frac{1960}{200}\)= 9.8 m/s

Question 7.
Does a body have acceleration in the following situations? Why?

  • body travelling along a straight line with uniform velocity.
  • body travelling along a straight line with non-uniform velocity.
  • body travelling along a circular path with uniform speed.
  • body travelling along a circular path with non-uniform speed.

Answer:

  • body travelling along a straight line with non-uniform velocity.
  • body travelling along a circular path with non-uniform speed.

If an object should have acceleration it should be travelled in non uniform speed or in non uniform velocity.
Change in velocity time
Acceleration = \(\frac{change in velocity}{time}\)

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Question 8.
A lorry travelling with a velocity of 30 m/s came to rest in 5 s. What is its acceleration?
Answer:
Acceleration = \(\frac{0 – 30}{5}\) = 6m/s²
∴ Retardation = 6m/s²

Question 9.
What is the displacement of a car in 30 s if it is travelling with a velocity of 15 m/s?
Answer:
Velocity = \(\frac{dispalcement}{time}\)
Displacement = velocity × time = 15 × 30 = 450m

Question 10.
Observe the figure showing the path of a body which started from
A and moved to C through B.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 5
a. Calculate the speed of the body.
b. What is the velocity of the body?
c. What is the velocity of the body if it has taken 5 s to reach A hack from C?
d. Compare the velocity of the body when it reached C from A and also when it reached A from C.
Answer:
a. Distance = 60 + 40 = 100 m
Time = 6 + 4 = 10 sec
∴ Speed = \(\frac{100}{10}\) = 10m/s

b. Velocity = \(\frac{dispalcement}{time}\) = \(\frac{70}{10}\) = 7m/s
c. displacement = 0, Velocity = 0
d. Velocity of the body when it reached C from A = 7 m/s
Velocity of the boby when it reached A from C velocity = 0 m/s

Motion Additional Questions and Answers

Question 1.
Observe the figure in which a child travelled from A to B and returned from B to A.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 6

a. Are the distance from A to B and B to A through C be equal?
b. What is the difference between the distance between A to B and return to B through C?
c. Which is the displacement ?
Answer:
a. No
b. The path from A to B is straight path. But the path from B to A is curved.
c. A to B

Question 2.
If a stone is thrown up 8 m perpendicularly and it falls in hand itself,
a. What is the distance travelled by the stone when it is thrown up and what is the displacement?
b. What is the distance when it fell in hand and what is the displacement?
Answer:
a. Distance = 8m, displacement = 8m
b. Distance = 16m, displacements = 0

Question 3.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 7
A man start from A travelled through B, C and D reaches at A. Complete the table below.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 8
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 9

Question 4.
Tabulate the difference between distance and displacement
Answer:

Distance displacement
Scalar quantity Vector quantity
Motion in different direction Straight line motion

Question 5.
A car travels a distance of 15m from A to B in 2 s, a distance of 25 m from B to C in 3 s, a distance of 8 m from C to D in 1 s.
a. What is the total distance travelled?
b. What is the total time taken to travel this distance?
c. What is the average speed of the car?
Answer:
a. 48 m
b. 6 Sec
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 12

Question 6.
Write examples of uniform velocity and non uniform velocity.
Answer:
mniform velocity : Travelling 9# light non uniform velocity : Travelling of bus

Question 7.
What is the acceleration when a car started from rest raised a velocity 80 m/s in 10 s?
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 16

Question 8.
Complete the table using the figure below.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 10

Answer:
Travelled position acceleration
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 11

Question 9.
Write 3 occasions of occurrence of acceleration.
Answer:
1. Falling of coconut from coconut tree.
2. Rolling of wheel on a slope.
3. A train starting from the station.

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Question 10.
write three methods to reduce road accidents.
Answer:
1. walk along the footpath only
2. Avoid over speed, Avoid drink and drive

Question 11.
Brakes are applied suddenly to a racing car travelling at 50m/s. If the car stops after 20 seconds, calculate the retardation of the car.
Answer:
Initial velocity of the car (u) = 50 m/s
Final velocity of the car (v) = o (The car stops)
Time = 20s
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 13

When it is written as retardation, no negative sign is required.
∴ Retardation of the car = 2.5 m/s²

Question 12.
Classify the following quantities into Scalar and Vector quantities.
Distance, Displacement, Time, Speed, Velocity, Acceleration.
Answer:

Scalar Vector
Distance Displacement
Time Velocity
Speed Acceleration

Question 13.
Velocity is a vector quantity. What do you mean by vector quantities? Write examples.
Answer:
Quantities having both magnitude and direction are called vector quatities. eg: Displacement, Acceleration.

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Question 14.
Compare distance and displacement.
Answer:
The length of the path covered by a body is known as distance. This is a scalar quantity. This is measured in a standard unit metre. The straight line distance between initial and final positions of a moving body is called displacement. This is a vector quantity. This is measured in metre.

Question 15.
A cheetah can start from rest and attain the velocity 72 km/h in 2 seconds. Calculate the acceleration of cheetah.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 14

Question 16.
Given figure shows the two ways in which a person reached at C
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 9 Motion 15

Reach at C from A through B ,
then
a. From fig.1 what is the total di-stance travelled? What is the total displacement covered?
b. From fig.2 what is the total distance travelled by the person? What is the total displacement covered by him?
c. find out the situation in which distance and displacement are equal by comparing given measures?
Answer:
a. Distance = 14 m
Displacement = 10 m
b. Distance = 14 m,
Displacement = 14 m
c. Only when a moving object travelling in a straight line on same direction.

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Plus One Computer Science Notes Chapter Wise HSSLive Kerala

HSE Kerala Board Syllabus HSSLive Plus One Computer Science Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala HSSLive Plus One Notes. Here HSSLive.Guru has given Higher Secondary Kerala Plus One Computer Science Chapter Wise Quick Revision Notes based on CBSE NCERT Syllabus.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus One
Subject Computer Science
Chapter All Chapters
Category Kerala Plus One

Kerala Plus One Computer Science Notes Chapter Wise

We hope the given HSE Kerala Board Syllabus HSSLive Plus One Computer Science Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus One Computer Science Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

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Plus Two Hindi Previous Year Question Papers and Answers Kerala

HSE Kerala Board Syllabus Plus Two Hindi Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both Hindi medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Hindi Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board SCERT, Kerala Board
Textbook NCERT Based
Class Plus Two
Subject Hindi
Papers Previous Papers, Model Papers, Sample Papers
Category Kerala Plus Two

Kerala Plus Two Hindi Previous Year Question Papers and Answers

We hope the given HSE Kerala Board Syllabus Plus Two Hindi Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both Hindi medium and Malayalam medium will help you. If you have any query regarding HSS Live Kerala Plus Two Hindi Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

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