Prasanna

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 3 Electrochemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 3 Electrochemistry

Question 1.
From the position of elements in the electrochemical series, copper (Cu) can displace silver (Ag) from silver nitrate solution. (March – 2010)
a) Represent the cell constructed with silver and copper electrodes.
b) Write down the reaction taking place at the anode.
c) Write down the reaction taking place at the cathode.
d) Write the Nernst equation for the above cell reaction.
Answer:

Question 2.
In a class room, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday’s laws of electrolysis. (May – 2013)

a) State the Faraday’s laws of electrolysis.
b) Explain the term electrochemical equivalent.
c) Calculate the quantity of electricity required to deposit 0.09 g of Aluminium during the following electrode reaction:
Al³⁺ + 3e → Al (Atomic mass of Al = 27)
Answer:
a) First law : The amount of a substance which is deposited or liberated at any electrode during ectrolysis is directly proportional to the quantity of electricity flowing through the electrolyte.

Second law: If the same quantity of electricity is passed through different electrolytes the amount of substances formed is directly proportional to their chemical equivalent weights.

b) It is the quantity of a substance formed when one-ampere current is passed through an electrolyte for one second.

c) Quantity of electricity required to deposit 27 g of
Al = 3F = 3 x 96500 C = 289500 C
∴ the quantity of electricity required to deposit 0.09
\(g \text { of } A l=\frac{289500 \times .09}{27}=965 C\)

Question 3.
The limiting molar conductivity of an electrolyte is to Aained by adding the limiting molar conductivities of cation and anion of the electrolyte. (March – 2011)
a) Name the above law.
b) What is meant by limiting molar conductivity?
c) Explain how conductivity measurements help to determine the ionization constant of a weak electrolyte like Acetic Acid.
d)Explain the change of conductivity and molar conductivity of a solution with dilution.
Answer:
a) Kohlrauschs law.
b) It is the conductivity of an electrolyte when the concentration of the solution approaches zero (or at infinite dilution).
c) The limiting molar conductivity of acetic acid \(\left(\Lambda_{\mathrm{Gh}_{3} \mathrm{COOH}}\right)\) isdeterrnined by app’ying Kohlrauschts law \(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{0}=\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}+\Lambda_{\mathrm{HCl}}^{0}-\Lambda_{\mathrm{NaCl}}^{0}\right)\). Then, degree of dissociaijon,OE is determined using the re lation, \(a=\frac{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{C}}}{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{O}}}\)

From α, the diissociation constant can be deter mined using the relation, \(K_{a}=\frac{c a^{2}}{(1-\alpha)}\)

Substituting for CL we get,

d. Conductivity decreases with dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution. The vanation of molar conductivity with dilution is different for strong and weak electrolytes. For strong electrolytes molar conductivity in creases steadily with increase in dilution due to decrease in interionic attraction. Thus, a straight line is obtained when ∧_m is plotted against C^1/2. For weak electrolytes molar conductivity increases with dilution steadily intially and shows a steep increase, especially at lower concentration due to increase in degree of dissociation.

Thus, a plot of ∧_m against C^1/2 gives a curve,

Question 4.
The standard electrode potentials of some electrodes are given below: (May – 2011)
E°_(zn2+,zn) = – 0.76V
E°_{(cu2+, Cn)} = + 0.34V
E°_{(Ag+, Ag)} = + 080V
E°_{(H+, H2)} = 0V

a) Can CuSO₄ solution be kept in silver vessel?
b) Zinc or Copper which can displace hydrogen from dii. H₂SO₄?
c) What is the reaction taking place at SHE when its connected to Ag/Ag electrode to form a galvanic cell?
d) Find the value of K_C (equillibnum constant) in the Daniel cell at 298k.
Answer:
a) Yes.
Since the standard reduction potential of silver is more than that of copper it is less reactive than copper and hence cannot react with CuSO₄.

b) Zinc with negative standard electrode potential is more active than hydrogen and hence can displace hydrogen from dil.H₂SO₄. But copper with a positive standard electrode potential is less active than hydrogen and hence cannot displace hydrogen from dil. H₂SO₄.

c) Since silver is less active than hydrogen, when silver electrode is connected with S.H.E, silver will act as cathode and S.H.E will act as anode. At the anode H is oxidiseci to H⁺. Hence, the reaction taking place at S.H.E is

Question 5.
Daniell cell is a galvanic cell made of zinc and copper electrodes. (March – 2012)
i) Write anode and cathode reactions in Daniell cell.
ii) Nernst equation for the electrode reaction
Mⁿ⁺ + + ne- 2 M is
\(\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}=\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}^{0}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{1}{\left[\mathrm{M}^{\mathrm{n}^{+}}\right]}\)
Derive Nernst equation for Daniel cell.
OR
Leclariche cell, Lead storage cell and Fuel cell are galvanic cells having different uses.

i) Among these, the Leclanche cell is a primary cell and Lead storage cell is a secondary cell. Write any two differences between primary cells and secondary cells.
ii) What is a Fuel cell?
iii) Write the overall cell reaction in H₂ – O₂ Fuel cell.
Answer:
i) Anode : Zn → Zⁿ²⁺ + 2e- (oxidation half reaction)
Cathode : Cu²⁺ + 2e^– → Cu (reduction half reaction)
Overall cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu Theceilcanberepresentedas: Znl Zn²⁺llCu²⁺lCu
Anode : Zn I ZnSO₄
Cathode : Cu I CuSO₄

which is the Nernst equation for Daniell cell.
OR

1) Primary cell	Secondary cell
a) Electrode reaction cannot be reversed.	a) Electrode reaction can be reversed by an external electric energy source.
b) Reaction occur only once & after use they become dead; not chargeable.	b) Reaction can occur many times in both directions Rechargeable

ii) Fuel cells are galvanic cells in which chemical energy from fuels like H₂, CO, CH₄ etc are converted to electrical energy.

Question 6.
Innumerable number of galvanic cells can be constructed on the pattern of Daniell cell by taking combination of different half cells. (May – 2012)
i) What is galvanic cell?
ii) Name the anode and cathode of the Daniell cell.
iii) Write the name of the half-cell represented by Pt_(S)/H_2(g)/H⁺_(aq).
iv) What is the potential of the above half-cell at all temperate res?
Answer:
i) It is a device for converting chemical energy released into electrical energy.
ii) Anode – Zn rod dipped in ZnSO₄
Cathode – Cu rod dipped on CauSO₄
iii) Standard Hydrogen Electrode (S.H.E)
iv) S.H.E is assigned a zero potential at all temperatures.

Question 7.
With decrease in concentration of an electrolytic solution, conductivity (K)decreases and molar conductivity (∧_m) increases. (March – 2013)
i) Write the equation showing the relationship between conductivity and molar conductivity.
ii) How will you account for the increase in molar conductivity with decrease in concentration?
iii) Limiting molar conductivity (L°_m) of a strong electrolyte can be determined by graphical extrapolation method. Suggest a method for the determination of limiting molar conducivity of a weak electrolyte, taking acetic acid (CH₃COOH) as example.
Answer:
i) Molar conductivity \(\left(\Lambda_{m}\right)=\frac{\kappa \times 1000}{M}\)
where M → Molanty and K → Conductivity
OR
\(\lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{C}}\) Concentration of where solution in mol/litre.

ii) Molar conductivity increase with decrease in con centration or increase in dilution as number of ions as well as mobility of ions increases with dilution.

a) For strong electrolytes, the number of ions do not increase appreciably on dilution and only mobility of ions increases due to decrease in interionic attraction.

∴ increases a little as shown in the above graph.

b) For weak electrolytes, the number of ions, as well as mobility of ions, increases on dilution. Hence, increases steeply on dilution, especially near lower concentrations as shown in graph.

iii) Using Kohlrauschs law

Thus the molar conductivity of CH3COQH at infinite dilution can be determined from the knowledge Of \(\lambda_{\mathrm{m}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)}^{0}, \lambda_{\mathrm{m}(\mathrm{HCl})}^{0}, \lambda_{\mathrm{m}(\mathrm{NaCl})}^{0}\)

Question 8.
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells. (May – 2013)
a) What is a galvanic cell?
b) Name the cathode and anode used in the Daniell cell.
c) Name the cell represented by Pt_(S), H_2(g)/H⁺_(aq).
d) According to convention what is the potential of the above cell at all temperatures?
e) Write the use of the above cell.
Answer:
a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell,
b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.
c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode.
d) According to convention, S.H.E is assigned a zero potential at all temperatures
e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

Question 9.
a) The cell reactìon in Daniell cell is Zn_(s) + CU²⁺_(aq) → Zn²⁺_(aq) + CU_(s) and Nernst equation for single electrode potential for general electrode reaction \(\mathrm{M}^{\mathrm{n}+}{ }_{(\mathrm{aq})}+\mathrm{ne}^{-} \longrightarrow \mathrm{M}_{(\mathrm{s})}\) is \(E_{M^{n+} M}=E_{M^{n+} / M}^{0}-\frac{2.303 R T}{n F} \log \frac{[M]}{\left[M^{n+}\right]}\) Derive Nernst equation for Daniell cell. (March – 2014)
b) Daniell cell is a primary cell while lead storage cell is a secondary cell. Write any one difference between primary cells and secondary cells.
Answer:
a) In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we can write, For Cathode:

b) In primary cells the reaction occurs only once and after use over a period of time cell becomes dead and cannot be reused again. Here the cell reaction is irreversible. e.g., dry cell Secondary cells after use can be recharged by passing current through them in the opposite direction so that they can be used again. Here the cell reaction is reversible. e.g., Lead storage cell.

Question 10.
Fuel cells are special types of galvanic cells. (May – 2014)
a) i) Whataregalvaniccells?
ii) Write any two advantages of fuel cells.
b) Write the electrode reactions is H₂ – O₂ fuel cells.
a) i) These are devices for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as electrical energy. e.g., Daniell oeil.
ii) 1) Fuel cells are pollution free.
2) Fuel cells are highly efficient (about 70%) compared to thermal plants (about 40%)
3) Fuel cells run continuously as long as the reactants are supplied. (any two)

b) At cathode:
O₂(g) + 2H₂O(l) + 4e → 4OH(aq)
At anode:
2H₂(g) + 4OH(aq) → 4H₂O(l)

Question 11.
You are supplied with the following substances: Copper rod, Zinc rod, Salt bridge, two glass bea kers, a piece of wire, 1 M CuSO₄ solution, 1 M ZnSO₄ solution. (March – 2015)
a) Represent the cell made using the above materials.
b) i) Write the Nemst equation for the above cell.
ii) Calculate the standard EMF of the cell if
\(\begin{array}{l}
E_{\left(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\right)}=-0.76 \mathrm{~V} \\
\mathrm{E}_{\left(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\right)}=+0.34 \mathrm{~V}
\end{array}\)
Answer:
a) Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s)
OR

Question 12.
a) Conductance (G), conductivity (K) and molar conductivity (∧_m) are terms used in electrolytic conduction. (May – 2015)
i) Write any two factors on which conductivity depends on.
ii) How do conductivity and molar conductivity vary with concentration of electrolytic solution?
b) Write any one difference between primary cell and secondary cell.
Answer:
a) i) 1. the nature of the electrolyte added
2. size of the ions produced and their solvation
3. the nature of the solvent and its viscosity
4. concentration of the electrolyte
5. temperature (any two factors)

ii) Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity, ∧_m = _kV

∧_m increases with decrease in concentration. This is because the total volume (V) of the solution containing one mole of electrolyte also increases. The decrease in K on dilution is more than compensated by increase in its volume.

In the case of strong electrolytes Am increases slowly with dilution and can be represented by the equation:
\(\Lambda_{m}=\Lambda_{m}^{0}-A c^{1 / 2}\)

where ‘c’ is the molar concentration, ‘A’ is a constant (equato to -ve of slope) and ∧_m° is the limiting molar conductivity. Here, the plot of ∧_m against ‘C^1/2‘ will be a straight line. In the case of weak electrolytes ∧_m increases steeply on dilution, especially near lower concentrations due to increase in degree of dissociation.

b) Pnmary cell – Cell in which the reaction occurs only once and after use over a period of time the cell becomes dead and cannot be reused again. Secondary cell – Cell which can be recharged after use by passing current through it in the opposite direction so that it can be used again.

Question 13.
a) Which of the following is a secondary cell? (March – 2016)
a) Dry cell
b) Leclanche cell
C) Mercury cell
d) None of these

b) What is the relationship between resistance and conductance?
c) One of the fuel cells uses the reaction of hydrogen and oxygen to form water. Write down the cell re action taking place in the anode and cathode of that fuel cell.
Answer:
d) None of these
b) Resistance is inversely proportional to conductance. Or, Conductance is the inverse of resistance.

Question 14.
Galvanic cells are classified into primary and secondary cells (May – 2016)
a) Write any two differences between primary cell and secondary cell.
b) i) What is a fuel cell?
ii) Write the overall cell reaction in H₂ – O₂ fuel cell.
Answer:
a) Primary cell
Cell reaction cannot be reversed and hence can not be recharged, cannot be reused again. e.g. Dry cell, Mercury cell

Secondary cell
Cell reaction can be reversed and hence can be recharged, can be resued again. e.g. Lead storage battery, nickel-cadmium cell

b) i) Fuel cell is a galvanic cell that is designed to convert the energy of combustion of fuels directly into electncal energy.
ii) 2H₂(g) + O₂(g) → 2H₂O(l)

Question 15.
a) Represent the galvanic cell based on the cell reaction given below (March – 2017)
\(\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}\)
b) Write the half cell reaction of the above cell.
c) ∧_m⁰ for NaCI. HCI and NaAc are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧_m⁰ for HAc.
Answer:

Question 16.
a) Identify the weak electrolyte from the following: (May – 2017)
i) KCl
ii) NaCl
iii) KBr
iv) CH₃COOH

b) Kohlrausch’s law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.
i) State Kohlrausch’s law.
ii) The molar conductivity ∧_m of .001 M acetic acid is 4.95 x 10^-5 S cm² mol^-1. Calculate the degree of dissociation (α) at this concentra tion if limiting molar conductivity \(\wedge_{m}^{0}\) for H+ is 340 x 10^-5 S cm² mol^-1 and for CH₃COO is 50.5 x 10^-5 S cm² mol^-1.
Answer:
a) iv) CN₃COOH
b) i) Kohlrausch’s law: Limiting molar conductivity of an electrolyte is the sum of individual contnbutions of anion and cation respectively.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 2 Solutions.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 2 Solutions

Question 1.
Colligative properties are properties of solutions which depend on the number of solute particles irrespective of their nature. (March – 2010)
a) Name the four important colligative properties.
b) What happens to the colligative properties when ethanoic acid is treated with benzene? Give reason.
Answer:
a) 1) Relating lowering of vapour pressure of the solvent
2) Depression of freezing point of the solvent
3) Elevation of Boiling point of the solvent
4) Osmotic pressure of the solution

b) Molecules of ethanoic add (acetic acid) dimenses in benzene due to hydrogen bonding. As a result of dimerisation, the actual number of solute particles in the solution is decreased. As colligative property decreases molecular mass increases.

Question 2.
a) Mr. Raju has determined the molecular masses of different solutes in different solvents by osmotic pressure measurements and presented them in the following table. Please help him to complete the table. (May – 2010)

Solute	Solvent	Theoretical Molecular Mass	Experimental Molecular Mass
NaCI Benzoic acid Urea Acetic acid CaCI2 Glucose Ai2(So4)3	Water Benzene Water Benzene Water Water Water	A B C D E F G	A/2 — — — — — —

b) The extent of deviation from ideal behaviour of a solution is explained by van’t Hoff factor, i. What is meant by van’t Hoff factor?
Answer:
a) 2B, C, 2D, E/3, F, G/5.
b) 2B and 2 D are due to association
\(\mathrm{i}=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}\)
OR
\(\mathrm{i}=\frac{\text { Observed colligative property }}{\text { Calculated colligative property }}\)

Question 3.
Colligative properties can be used to determine the molecular mass of solutes in solutions. (March – 2011)
a) What do you mean by ‘Colligative Property’?
b) Fordeterminingthe molecular mass of polymers, osmotic pressure is preferred to other properties. Why?
c) For intravenous injections only solutions with freezing point depression equal to that of 0.9% NaCI solution is used. Why?
Answer:
a) Colligative properties are those properties which depend upon no. of particles in the solution.

b) i) Pressure measurement can be done around the room temperature.
ii) Molarity of the solution is used instead of molality.
iii) Its magnitude is large compared to other colligative properties.
iv) Polymers have poor solubility.

c) This is because the osmotic pressure associated with the fulid inside the blood cell is equivalent to that of 0.9 (m/w) sodium chloride solution, i.e., blood is isotonic with this solution.

Question 4.
Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure are important colligative properties of dilute solutions. (May – 2011)
a) Relative lowering of vapour pressure of an aqueous dilute solution of glucose is 0.018. What is the mole fraction of glucose in the solution?
b) An aqueous dilute solution of a non-volatile solute boils at 373.052 K. Find the freezing point of the solution.
For water K_b = 0.52K Kg mol^-1
For water K_f = 1.86 K Kgmol^-1
Normal boiling point of water = 373K
Normal freezing point of water = 273K
Answer:
a) 0.018
Because according to Raoult’s law for solutions containing non-volatile solutes, the relaive lower-ing of vapour pressure is equal to the mole fraction of the non-volatile solute.

Question 5.
Vapour pressure of a solution is different from that of pure solvent. (March – 2012)
i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
ii) State the above law.
iii) Vapour pressure of chloroform (CHCI₃) and dichloro methane (CH₂CI₂) at 298 K are 200 mm and 415 mm of Hg respectively. Calculate the vapour pressure of solution prepared by mixing 24 g of chloroform and 17 g of dichloro methane at 298 K. [At. Mass : H -1, C – 12, Cl – 35.5]
Answer:
i) Raoult’s law.

ii) It states that at a given temperature for a solution of volatile liquids, the partial v.p. of each component in the solution is directly proportional to its mole fraction.

Question 6.
Colligative properties are properties of solutions which depend on the number of solute particles in the solution. (May – 2012)
i) Write the names of four important colligative properties.
ii) The value of van’t Hoff factor, ‘i’ for aqueous KCI solution is close to 2, while the value of ‘i’ for ethanoic acid in benzene is nearly 0.5. Give reason.
i) The Colligative Properties are:
1) Relative lowering of vapour pressure
2) Elevation of boiling point
3) Depression of freesing point
4) Osmotic pressure

ii) This is caused by dissociation in the case of KCI and association in the case of acetic acid.

KCI in aqueous solution undergoes dissociation as KCI → K⁺ + Cl^–

Thus, if complete ionisation occurs the number of particles in solution becomes double and hence van’t Hoff factor (i) for aqueous KCI solution is close to 2.

In the case of ethanoic acid (acetic acid) association (dimerisation) occurs in benzene through in- termolecular hydrogen bonding. Thus, if complete association occurs the number of particles in solution becomes half and hence van’t Hoff factor (i) for ethanoic acid in benzene is nearly 0.5.

Question 7.
Elevation of boiling point is a colligative property, (March – 2013)
i) What are colligative properties?
ii) Elevation of boiling point (D T_b) is directly proportional to molality (m) of solution.
Thus, DT_b = K_bm, K_b is called the molal elevation constant.
From the above relation derive an expression to obtain molar mass of the solute,
iii) The boiling point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of the solute. K_b for benzene is 2.53K kg mol^-1.
Answer:
i) Colligative properties are those properties which depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

Question 8.
Liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. (May – 2013)
a) State Raoult’s law.
b) What are ideal solutions?
c) Write two important properties of ideal solutions.
d) What type of deviation is shown by a mixture of chloroform and acetone? Give reason.
Answer:
a) Raoult’s law states that for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction. Or The partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of pure component and mole fraction of that component in the solution.

b) Ideal solutions are solutions which obey Raoult’s law over the entire range of concentration.

c) i) Enthalpy of mixing of the pure components to form the solution is zero, i.e.; Δ_mixH = 0
ii) Volume of mixing is zero, i.e.; Δ_mixV = 0

d) Negative deviation. This is because chloroform molecule is able to form hydrogen bond with acetone molecule.

Question 9.
Osmotic pressure is a colligative property and it is proportional to the molarity of the solution. (March – 2014)
a) What is osmotic pressure?
b) Molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.
c) Calculate the osmotic pressure exerted by a solution prepared by dissolving 1.5 g of a polymer of molar mass 185000 in 500 mL of water at 37°C. [R = 0.0821 L atm K^-1 mol^-1].
Answer:
a) Osmotic pressure is the extra pressure applied on the solution to stop osmosis
b) NaCI, as a strong electrolyte dissociates to form Na+ and Ch ions. The number of particles doubles colligative property also doubles. Observed molecular mass will be half.

Question 10.
Molarity (M), molality (m) and mole fraction (x) are some methods for expressing concentrations of solutions. (May – 2014)
a) Which of these are temperature independent?
b) i) Define ‘molefraction’.
ii) A mixture contains 3.2 g methanol (molecular mass = 32 u) and 4.6 g ethanol, (molecular mass = 46 u) Find the molefraction of each ’ component)
Answer:
a) Molality and Mole fraction are temperature independent because these are mass to mass relationships. Mass is independent of temperature,
b) i) Mole fraction is defined as the ratio of the number of moles of that component to the total number of moles of all the components in solution.

Question 11.
a) Among the following which is not a colligative property? (March – 2015)
i) Osmotic pressure
ii) Elevation of boiling point
iii) Vapour pressure
iv) Depression of freezing point

b) i) 200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of solution at 300 K is found to be 8.3 x 10^-2 bar. Calculate the molar mass of protein. (R = 0.083 L bar K^-1 mol^-1)
ii) What is the significance of van’t Hoff factor?
Answer:
a) iii) Vapour pressure
b) i) Osmotic pressure, n = 8.3 x 10^-2 bar
Volume of the solution, V = 200 cm³ = 0.200 L
Temperature, T = 300 K
R = 0.083 L bar mol^-1 K^-1

Thus, in the case of association, the value of ‘i’ is less than unity while for dissociation ‘i’ is greater than unity. If i =1 it means that there is no association or dissociation of the solute particles in solution.

Question 12.
a) Draw a vapour pressure curve, by plotting vapour pressure against mole fraction of an ideal solution of two volatile components A and B (not to scale). Indicate partial vapour pressure of A and B (P_A and P_B) and total vapour pressure (P_total) (May – 2015)
b) What is an ideal solution?
c) Modify the above plot for non-ideal solution showing positive deviation. (Draw the above plot once again and modify).
Answer:

b) A solution which obeys Raoult’s law over the entire range of concentration is known as an ideal solution.

Question 13.
a) Number of moles of the solute per kilogram of the solvent is (March – 2016)
a) Mole fraction
b) Molality
c) Molarity
d) Molar mass

b) The extent to which a solute is dissociated or associated can be expressed by Van’t Hoff factor.’ Substantiate the statement.
c) The vapour pressure of pure benzene at a certain temperature is0.850 bar. Anon volatile, non-electrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g mol^-1), the vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?
Answer:
a) b) Molality
b) b) The van’t Holf factor,

When i = 1 ⇒ there is no association or dissociation of solute particles.
When i < 1 ⇒ there is association of solute particles. When i > 1 ⇒ there is dissociation of solute particles.

Question 15.
Osmotic pressure is a colligative property. (May – 2016)
a) What is osmotic pressure?
b) 1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 k. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of solute.
Answer:
a) Osmotic pressure is the extra pressure applied on the solution side to just stop osmosis i.e., the flow of the solvent from its side to solution side through the semipermeable membrane,

Question 16.
a) Henry’s law is related to solubility of a gas in liquid. (March – 2017)
i) State Henry’s law.
ii) Write any two applications of Henry’s law.
b) 1000cm³ of an aqueous solution of a protein contains 1.26 gm of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 x 10^-3 bar. Calculate molar mass of the protein. (R = 0.083 L bar mol^-1 K^-1)
Answer:
a) i) Henry’s law states that at constant tempera-ture, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = K_H x where, p is the partial pressure of the gas in vapour phase, K_H is the Henry’s law constant and x is the mole fraction of the gas in the solution.

ii) 1. To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

2. To avoid bends and the toxic effects of high concenration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium.

3. Low partial pressure of oxygen at high altitudes leads to low concentration of oxygen in the blood and tissues of people living at high altitudes or climbers and causes anoxia. (Any two applications)

Question 17.
a) The mole fraction of water in a mixture containing equal number of moles of water and ethanol is (May – 2017)
i) 1
ii) 0.5
iii) 2
iv) 0.25

b) The following are the vapour pressure curves of a pure solvent and a solution of a non-volatile solute in it.

Based on the above curves answer the following questions:
i) What do the curves A and B indicate?
ii) Explain why the value of TB is greater than that of Tb⁰.
Answer:
a) ii) 0.5
b) i) A-Vapour pressure curve of solvent Vapour pressure curve of solution
ii) Due to the presence of a non-volatile solute vapour pressure of solution is less than solvent and the boiling point is increased.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 1 The Solid State.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 1 The Solid State

Plus Two Chemistry The Solid State 4 Marks Important Questions

Question 1
a) Schottky defects and Frenkel defects are two stoichiometric defects shown by crystals. (March – 2010)
i) Classify the following crystals into those showing Schottky defects and Frenkel defects:
NaCI, AgCI, CsCI, CdCI₂
ii) Name a crystal showing both Schottky defect and Frenkel defect.
b) Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.
i) ↑↓↑↓↑↓↑↓
ii) ↑↑↓↑↓↑↑↑
iii) ↑↑↑↑↑↑↑↑
Answer:
a) i) Schottchydefect – NaCI, CsCI Frenkel defect – AgCI, CdCI₂
ii) AgBr

b) i) Antifero magnetism
ii) Ferrimagnetism
iii) Ferromagnetism

Question 2.
Based on the nature of order present in the arrangement of the constituent particles, solids are classified into two, crystalline and amorphous. (May – 2010)
a) List out any four points of difference between crystalline and amorphous solids.
b) A list of solids are given below:
Quartz, glass, iodine, ice.
From this, identify crystal (s)
i) having sharp melting point.
ii) which is/are isotropic
Answer:

Crystalline	Amorphous
i) Long-range order ii) Sharp melting point iii) Newly formed surface is smooth iv)  Anisotropic	i) Short-range order ii) Range of melting point iii) Newly formed surface is rough. iv) Isotropic

b) i) Quartz, Iodine, Ice
ii) Glass

Question 3.
Cristal defects give rise to certain special properties in the solids. (March – 2011)
a) What is meant by Frenkel Defect?
b) Why does LiCI not exhibit Frenkel Defect?
c) Explain the pink colour of LiCI when heated in . the vapours of Li.
Answer:
a) The dislocation of a cation from its original site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
b) The size of the cation is bigger than the void.
c) Due to F – centre. It is an electron trapped anion vacancy.

Question 4.
A cubic unit cell is characterized by a = b = c and α = β = γ = 90° (May – 2011)
a) Name three important types of cubic unit cells and calculate the number of atoms in one unit cell in the above three cases.
b) A metal forms cubic crystals. The mass of one unit cell of it is M/NA gram, where M is the atomic mass of the metal and NA is Avogardo Number. What is the type of cubic unit cell possessed by the metal?
Answer:
a) Simple cubic unit cell or Primitive unit cell, Body – centred cubic unit cell (bcc) and Face – centred cubic unit cell (fee).
b) Primitive cubic unit cell: This unit cell has atoms only at its comers. There are 8 corners for a cube.
Contribution by atom at the corner = 1/8
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

Body – centred cubic unit cell:
This unit cell has an atom at each of its corners and also one atom at its body centre.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

Face – centred cubic unit cell:
This unit cell contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom Contribution by an atom at the face centre = \(\frac{1}{2}\)

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms

∴ Total number of atoms per unit cell = 4 atoms

c) Mass of one unit cell = \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}} \mathrm{g}\)

Mass of 1 mole of unit cells \(\mathrm{N}_{\mathrm{A}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) = M gram = Gram atomic mass It means that 1 unit cell contains one atom of the metal. Hence, the type of unit cell is primitive cubic or simple cubic.

Question 5.
Solids can be classified into three types on the basis of their electrical conductivities. (March – 2012)
i) Name three types of solids classified on the basis of electrical conductivities.
ii) How will you explain such classification based on Band theory?
Answer:
i) Conductors, insulators & semiconductors
ii) In conductors, the valance band overlaps with Ir e conduction band or no energy gap exists between the valance band and conduction band.

∴ The electrons can easily go into the conduction band and hence metals are good conductors. In insulators, the energy gap between valance band and conduction band is very large. Hence electrons from valance band cannot move into the conduction band.

Semi conductors have energy gap between conductors and insulators. At room temperature, these are not good conductors. But with an in crease in temperature electrons acquire sufficient energy to move from valance band into conduction band resulting in an increase in conductivity.

Question 6.
Schottky and Erenkel defects are stoichiometric defects. (May – 2012)
i) Write any two differences between Schottky defect and Frenkel defect.
ii) When pure NaCI (Sodium Chloride) crystal is heated in an atmosphere of sodium vapours, it turns yellow. Give reason.
Answer:
i)

Schottky defect	Frenkel defect
1. Vacancy defect which arises due to the messing of equal number of cations and anions from the lattice sites.	Interstitial defector dislocation defect which arises when the smaller ion, usually cation is dislocated from its normal site to an interstitial site.
3. Cation and anion in are of almost similar sizes	there is a large difference size of ions
4. The density of the crystal is lowered	It does not affect the density of the crystal

ii) It is caused by metal excess defect due to anion vacancies. When crystals of NaCI are heated in a atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCI. The electrons released from Na atoms diffuse into the crystal and occupy anionic sites to form Fcentres, which impart yellow colourto the crystal. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 7.
a) NaCI has fcc structure. Calculate the number of NaCI units in a unit cell of NaCI. (March – 2013)
b) Calculate the density of NaCl, if edge length
of NaCI unit cells is 564pm. [Molar mass of
NaCI =58.5g/mol].
Answer:
a) Number of Na ions = 12 (at edge centres) x \(\frac{1}{4}\) +
= 1(at body centre) x 1
= 3 + 1 = 4
Number of Cl- ions = 8 (at the corners) x \(\frac{1}{8}\) +
= 6 (at face centres) x \(\frac{1}{2}\)
= 1 + 3 = 4
∴ Number of NaCI units per unit cell (z) = 4

Question 8.
Unit cells can be broadly classified into 2 categories primitive and centred unit cells. (May – 2013)
a) What is a unit cell?
b) Name the three types of centred unit cells.
c) The unit cell dimension of a particular crystal system is a = b = c, α = β = γ = 90°. ldentify the crystal system.
d) Give one example for the above crystal system.
Answer:
a) The smallest repeating unit of a crystal.
b) Body centred unit cell, Face centred unit cell and End centred unit cell.
c) Cubic crystal system.
d) NaCI (Rock salt struãture)

Question 9.
a) Every substance has some magnetic properties associated with it. How will you account for the following magnetic properties? (March – 2014)
i) Paramagnetic property
ii) Ferromagnetic property

b) A compound is formed by two elements P and Q. Atoms of Q (as anions) make hep lattice and those of the element P (as cations) occupy all the tetrahedral voids. What is the formula of the compound?
Answer:
a) i) Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired . electrons, e g., O₂, Cu²⁺

ii) Ferromagnetic substances are strongly attracted by a magnetic field. They are permanantly magnetised. When a ferromagnetic substance is placed in a magnetic field all the magnetic domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced, e.g,, Fe, Co

For hep lattice, No. of particles per unit cell = 6

∴ Number of anions of Q in the unit cell = 6
Number of tetrahedral voids = 2 x N = 2 x 6 = 12
∴ Number of cations of P in the unit cell = 12
Hence, formula of the compound = P₁₂Q₂ = P₂Q

Question 10.
a) Crystalline solids are ‘anisotropic’. What is ‘anisotropy’? (May – 2014)
b) Copper crystals have fee unit cells.
i) Compute the number of atoms per unit cell of copper crystals.
ii) Calculate the mass of a unit cell of copper crystals. (Atomic mass of copper = 63.54 u)
Answer:
a) Anisotropy means physical properties shows different values along different directions, eg. refrative index electrical resistance,
b) i) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners x \(\frac{1}{8}\) per corner atom

Question 11.
Unit cells can be divided into two categories, primitive and centred unit cells. (March – 2015)
a) Differentiate between Unit Cell and Crystal Lattice.
b) Calculate the number of atoms per unit cell in the following:
i) Body centred cubic unit cell (bcc)
ii) Face centred cubic unit cell (fee)
Answer:
a) Unit Cell – It is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

Crystal Lattice – It is the regularthree dimensional arrangement of points in space.

b) i) Body centred cubic unit cell (bcc) – It has atoms at each of its corners and one atom at its body centre.
8 corners x \(\frac{1}{8}\) per corner atom = 8 x \(\frac{1}{8}\) = 1 atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

ii) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.

8 corners x \(\frac{1}{8}\) per corner atom
\(=8 \times \frac{1}{8}=1\) atom

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms
∴ Total number of atoms per unit cell = 4 atoms

Question 12.
a) Which of the following is not a characteristic of a crystalline solid? (May – 2015)
i) Definite heat of fusion
ii) Isotropic nature
iii) A regular ordered arrangement of constituent particles
iv) A true solid

b) Frenkel defect and Shottky defects are two stoichiometric defects found in crystalline solids.
i) What are stoichiometric defects?
ii) Write any two differences between Frenkel defect and Schottky defect.
Answer:
a) ii) Isotropic nature
b) i) Stoichiometric defects are those point defects which do not disturb the stoichiometry of the solid.

ii)

Schottky defect	Frenkel defect
1. Cation and anion in are of almost similar sizes	there is a large difference size of ions
2. The density of the crystal is lowered	It does not affect the density of the crystal

Question 13.
a) Which of the following is a molecular solid? (March – 2016)
a) Diamond
b) Graphite
c) Ice
d) Quartz

b) Unit cells can be classified into primitive and centered unit cells. Differentiate between primitive and centered unit cells.
c) Presence of excess Sodium makes NaCI crystal coloured. Explain on the basis of crystal defects.
Answer:
a) Ice
b) In primitive unit cell constituent particles are present only at the corner positions. Unit cells in which one constituent particles are present at the centres of a faces in addition to those at corners.

c) Such anionic sites occupied by unpaired electrons are called F – centres (colour centres). They impart yellow colourto the crystals of NaCI. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 14.
A unit cell is a term related to crystal structure. (May – 2016)
a) What do you mean by unit cell?
b) Name any two types of cubic unit cells.
c) Calculate the number of atoms in each of the above – mentioned cubic unit cells.
d) Identify the substance which shows Frenkel defect:
i) NaCI
ii) KCI
iii) ZnS
iv) AgBr
Answer:
a) Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
b) Simple cubic unit cell, body centred cubic (bcc) unit cell, face centred cubic (fee) unit cell (any two)
c) Number of atoms per unit cell
i) Simple cubic unit cell:
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

ii) Body centred cubic (bcc) unit cell: Total number of atoms in one unit cell
\(=\left(8 \times \frac{1}{8}\right)+(1 \times 1)=1+1=2\) atoms

iii) Face centred cubic (fee) unit cell: Total number of atoms in one unit cell \(=\left(8 \times \frac{1}{8}\right)+\left(6 \times \frac{1}{2}\right)=1+3=4\)
(any two required)

d) ZnS or AgBr
(AgBr shows both Schottky and Frenkel defects)

Question 15.
a) Identify the non – stoichiometric defect (March – 2017)
i) Schottky defect
ii) Frenkel defect
iii) Interstitial defect
iv) Metal deficiency defect
b) What type of substance could make better permanent magnets – ferromagnetic or ferrimagnetic? Justify your answer.
c) In terms of Band theory write the differences between conductor and insulator.
Answer:
a) iv) Metal deficiency defect

b) Ferromagnetic substances could make better permanent magnets because when these substances are placed in a magnetic field all the magnetic moments (domains) get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagentic substance becomes a permanent magnet.

The schematic alignment of magnetic moments in a ferromagnetic substance is as shown below:

c) In conductors the valence band is either partially filled or it is overlaped with a higher energy unoccupied conduction band so that the electrons can flow easily under an applied electric field. Whereas in insulators the energy gap between the filled valence band and the next higher unoccupied conduction band is large so that electrons cannot jump to it.

Question 16.
a) From the following choose the incorrect statement about crystalline solids. (May – 2017)
i) Melt at sharp temperature.
ii) They have definite heat of fusion.
iii) They are isotropic
iv) They have long range order.

b) Cubic unit cells are divided into primitive, bcc and fee.
i) Calculate the number of atoms in a unit cell of each of the following:
* bcc
* fcc

ii) Write two examples for covalent solids.
a) iii) They are isotropic
b) i) \(\begin{array}{l}
\text { bcc }-2\left(8 \times \frac{1}{8}+1=2\right) \\
\text { fCc }-4\left(8 \times \frac{1}{8}+6 \times \frac{1}{2}=4\right)
\end{array}\)
ii) Graphite, Diamond