Class 10 Chemistry Chapter 4 Notes Kerala Syllabus Gas Laws and Mole Concept Questions and Answers

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SSLC Chemistry Chapter 4 Notes Questions and Answers Pdf Gas Laws and Mole Concept

SCERT Class 10 Chemistry Chapter 4 Gas Laws and Mole Concept Notes Pdf

SSLC Chemistry Chapter 4 Questions and Answers – Let Us Assess

Question 1.
Convert the units of the given temperatures.

Answer:

°C	K
0	273
100	373
30	303
27	300
40	313

Question 2.
Complete the table. (Atomic Mass – H = 1, C = 12, O = 16, N = 14)

Answer:

Substance	Molecular mass	Given mass	Number of moles	Volume at STP
H2	2 g/mol	10g	5 moles	112L
CO2	44 g/mol	440g	10 moles	224L
NH3	17 g/mol	340g	20 moles	448L

Question 3.
A balanced equation of the reaction to produce ammonia is given.
N₂ + 3H₂ → 2NH₃
a) How many moles of hydrogen are required for 10 moles of nitrogen to react completely?
b) How much ammonia will be produced if 10 moles of nitrogen react completely?
Answer:
a) N₂ + 3H₂ → 2 NH₃
1 mole of N₂ reacts with 3 moles of hydrogen H₂ to form 2 moles of NH₃
Therefore, for 10 moles of nitrogen to react completely,

10 moles of N₂ × \(\frac{3 \text { moles of Hydrogen }}{1 \text { mole of Nitrogen }}\) 30 moles of H₂
So, 30 moles of hydrogen are required.

b) If 10 moles of nitrogen react completely, the amount of ammonia produced will be:
10 moles of N₂ × \(\frac{2 \text { moles of Ammonia }}{1 \text { mole of Nitrogen }}\) = 20 moles of NH₃
So, 20 moles of ammonia is produced

Question 4.
448 L of gas is stored in a cylinder at 0°C and 1 atm pressure.
a) How many moles of molecules does this gas contain?
b) What is the number of molecules in this sample?
Answer:
a) Volume of the gas (V) = 448 L
Temperature (T) = 0 °C (which is standard temperature)
Pressure (P) = 1 atm (which is standard pressure)
Number of moles = \(\frac{\text { Volume at STP }}{22.4 \mathrm{~L}}\) = \(\frac{448 \mathrm{~L}}{22.4 \mathrm{~L}}\)
= 20 moles
Therefore, the gas contains 20 moles of molecules,

b) Number ofmolecules = 20 × 6.022 × 10²³ molecules.

Question 5.
400 L of gas is stored in a cylinder at 27 °C and constant pressure.
a) What will be the temperature if the volume of this gas is reduced to 200 L at the same pressure?
b) Which gas law is relevant to this context?
c) The boiling point of a substance is 3 °C. Above what temperature in Kelvin does this substance obey the gas laws?
Answer:
a) Initial volume (V₁) = 400 L
Initial temperature (T₁ = 27 °C
Final volume (V₂) = 200 L
Pressure is constant.
T₁(K) = T₁(°C) + 273 = 27 + 273 = 300K
V₁/T₁ = V₂/T₂
T₂ = \(\frac{V_{2 \times} T_1}{V_1}\)
T₂ = \(\frac{200 \mathrm{~L} \times 300}{400 \mathrm{~L}}\)
T₂ = 150K

b) The gas law relevant to this context is Charles’s Law. This law describes the relationship between the volume and temperature of a gas at constant pressure.

c) Boiling point = 3 °C
Boiling point in Kelvin = 3 + 273 = 276 K. For the substance to obey the gas laws, it must be in the gaseous state. This occurs at temperatures above its boiling point.
Therefore, the substance obeys the gas laws above 276 K.

Question 6.
a) What is the number of Cl2 molecules in 710 g of chlorine gas?
b) What is the total number of atoms in this sample?
Answer:
a) Mass of chlorine gas = 710 g
Molar mass of Ch = 2 × 35.5 g/mol = 71 g/mol
Number of moles = 710 g / 71 g/mol = 10 moles
Number of Ch molecules = 10 moles × 6.022 × 10²³ molecules/mol
Therefore, there are 6.022 × 10²⁴ Cl₂ molecules in 710 g of chlorine gas.

b) Total number of atoms = Number of CI2 molecules × 2
Total number of atoms = 6.022 × 10²⁴ molecules × 2 atoms/molecule
Total number of atoms = 1.2044 × 10²⁵ atoms

Question 7.
How many grams of hydrogen are required to react with 700 g of nitrogen in the production of ammonia?
Answer:
N₂ + 3H₂ → 2NH₃
Molar mass of N₂ = 28g mol
Number of moles of N₂ = Mass of N₂ / Molar mass of N₂
Number of moles of N₂ = 700 g / 28g/mol = 25 moles
1 mole of N₂ reacts with 3 moles of H₂
Moles of H₂ required = 3 × moles of N₂
Moles of H₂ required = 3 × 25moles = 75 moles
Molar mass of H₂ = 2
Mass of H₂ required = Number of moles of H₂ × Molar mass of H₂
Mass of H₂ required = 75 moles × 2 g/mol = 150g

Question 8.
Calculate the following.
a) How many moles of CaCO₃ are present in its 1 kg?
b) Mass of the same number of NH₃ molecules as contained in 88 g of CO₂
c) Mass of 22.4 L CO₂ kept at STP.
(Hint: Atomic mass – H = 1, C = 12, O = 16, N = 14, Ca = 40)
Answer:
a) Molar mass of CaC0₃ = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol 1 kg = 1000 g
Number of moles = Mass / Molar mass
Number of moles of CaCO₃ = 1000 g / 100 g/mol = 10 moles
Therefore, there are 10 moles of CaCO₃ present in 1 kg.

b) Number of moles of CO₂ = Mass / Molar mass = 88 g / 44 g/mol = 2 moles
Molar mass of NH₃ = 14 + (3 × 1) = 14 + 3 = 17 g/mol
Mass of NH₃ = 2 moles × 17 g/mol = 34 g
Therefore, the mass of the same number of NH₃ molecules as contained in 88 g of CO₂ is 34 g.

c) 44g

Question 9.
The volume of a cylinder containing NH₃ at STP is 4480 mL.
a) Find the number of moles of NH₃ in the cylinder.
b) Find the mass of this NH₃ gas.
c) How many molecules of NH₃ are there in this gas cylinder?
Answer:
a) Number of moles = Volume / Molar volume at STP
Volume of NH₃ gas at STP = 4480 mL = 4.48 L
Number of moles = Volume / Molar volume at STP
Number of moles of NH₃ = 4.48 L / 22.4 L/mol = 0.2 moles
Therefore, there are 0.2 moles of NH₃ in the cylinder.

b) Molar mass of NH₃ = 14 + (3 × 1) = 14 + 3 = 17 g/mol
Mass of NH₃ = Number of moles × Molar mass = 0.2 moles × 17 g/mol = 3.4 g
Therefore, the mass of the NH₃ gas in the cylinder is 3.4 g.

c) Avogadro’s number = 6.022 × 10²³
Number of molecules = Number of moles × Avogadro’s number
Number of NH₃ molecules = 0.2 moles × 6.022 × 10²³ molecules/mol
Number of NH₃ molecules = 1.2044 × 10²³ molecules
Therefore, there are approximately 1.2044 × 10²³ molecules of NH₃ in the gas cylinder.

Question 10.
Consider the chemical reaction
2H₂ + O₂ → 2H₂O
a) How many moles of oxygen (O₂) should react to obtaining 10 mole H₂O?
b) Find the mass of oxygen gas (O₂) required to obtain 10 moles of H₂? (Atomic mass- H = 1, O = 16)
Answer:
a) 2H₂ + O₂ → 2H₂O
1 mole of O₂ produces 2 moles of H₂O. To get 10 moles of H₂O

Moles of O₂ = 10 moles H₂O × \(\frac{1 \text { mole of } \mathrm{O}_2}{2 \text { moles of } \mathrm{H}_2 \mathrm{O}}\) = 5 moles O₂
Therefore, 5 moles of oxygen (O₂) should react.

b) Molar mass of O₂ = 2 × 16 g/mol = 32 g/mol
Mass = Number of moles × Molar mass
Mass of O₂ = 5 moles × 32 g/mol = 160 g
Therefore, the mass of oxygen gas (O₂) required is 160 g.

Chemistry Class 10 Chapter 4 Notes Kerala Syllabus Gas Laws and Mole Concept

Question 1.
Why do these balloons rise in the air, float and come down?
Answer:
Helium gas is used to fill these balloons

Question 2.
Why do balloons filled with helium gas rise in the air?
Answer:
Helium balloons rise due to differences in density. Helium is much less dense than the air around it.

Question 3.
Is the density of helium gas filled in these balloons greater or lesser than that of the atmospheric air? Answer:
The density of helium gas in balloons is less than that of atmospheric air. This is why helium balloons can float in the air.

Question 4.
Tabulate the substances around as based on their physical states

Answer:

Solid	Liquid	Gas
Rock	Water	Air
Wood	Milk	Oxygen
Metal	Juice	Nitrogen
Ice	Oil	Steam
Sugar	Petrol	Helium

Question 5.
Do you know which of these states has the least density
Answer:
Gases are generally less dense than liquids and solids. This is because the atoms or molecules in a gas are much farther apart than in liquids or solids, meaning there is less matter packed into a given volume.

Question 6.
Find out the elements that exist in a gaseous state at ordinary temperature (20°C-25°C) and write them down in the science diary.
Answer:
Oxygen (O₂), Nitrogen (N₂), Hydrogen (H₂), Helium (He), Neon (Ne), Argon (Ar). These elements are gases at ordinary temperature.

Question 7.
From the following, tabulate the compounds that exist in a gaseous state at ordinary temperature.
[Carbon dioxide, sodium chloride, ammonia, glucose, ammonium chloride, methane, sulphur dioxide, carbon monoxide, Nitric acid, nitrogen dioxide.]
Answer:
Carbon dioxide (CO₂), Ammonia (NH₃), Methane (CH₄), Sulphur dioxide (SO₂), Carbon monoxide (CO), Nitrogen dioxide (NO₂).

Question 8.
Describe the arrangement of particles in a gaseous state.
(Compare the distance between the molecules, the force of attraction between them, the freedom of movement of the molecules and their energy in a gaseous state with those in the other states of matter and write them down.)


Answer:

Distance between molecules	Very high
Force of attraction between the molecules	Very low
Freedom of movement of the molecules	Very high
Energy of the molecules	Very high

Question 9.
If a liquid filled in a bottle of volume 1 L is transferred to a bottle of volume 2 L, what will be its volume?
Answer:
1 L

Question 10.
If oxygen gas filled in a bottle of volume 1 L is transferred to a bottle of volume 2 L, what will its volume be?
Answer:
2 L

Question 11.
What if a gas of volume 2 L is transferred to a bottle of 10 L volume?
Answer:
10 L

The volume of a gas is the volume of the container in which it is occupied.

Question 12.
Which molecules acquire energy due to their movement?
(Kinetic energy / potential energy)
Answer:
Kinetic energy

Question 13.
What happens to the kinetic energy of the molecules if the gas is heated? (Increases/decreases)
Answer:
Increases
When gases are heated, the energy of molecules increases, and hence, the temperature also increases.

Temperature is proportional to the average kinetic energy of gas molecules.

Question 14.
What happens to the pressure inside the syringe when the piston is pushed in? (Increases / Decreases) Answer:
The pressure inside the syringe increases when the piston is pushed in.

Question 15.
What happens when the piston is pulled back?
Answer:
The pressure inside the syringe decreases when the piston is pulled back.
A definite mass of a gas taken in a cylinder with a freely movable piston is illustrated in the figure.

Let us increase the pressure applied on this without changing the temperature

P(atm)	V(L)
2	10
4	5
10	2

Question 16.
What happens to the volume when pressure is increased in each situation?
Answer:
As pressure increases, volume decreases

Boyle’s law states that at a constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. That is, V ∝ l/p (temperature, mass constant) PV = k(k = a constant) (The value of k depends on the gas and its mass.

This relation was proposed by the scientist Robert Boyle in 1662.
If the volume of a fixed mass of gas is V₁ at pressure P₁ and V₂ at pressure P₂.
Then,

P1V1 = P2V2

Pressure – volume graph
To substantiate Boyle’s law, we can illustrate the relation between volume and pressure of a fixed mass of gas, using a graph. From this graph, we can see that PV is a constant at a constant temperature for a definite mass of gas.

PV is constant at all points in the graph.

Question 17.
Explain the following situations based on Boyle’s law.
a. The size of weather balloons increases as they rise above sea level.
b. The size of the air bubbles rising from the bottom of an aquarium increases as they reach the surface of the water.
Answer:
a) As it rises higher, atmospheric pressure decreases. Therefore, the volume of gas inside the balloon increases.
b) The pressure of the liquid is higher at the bottom of the aquarium. As it rises upwards, the pressure decreases; therefore, the volume of the air bubbles increases

Question 18.
The volume of a definite mass of gas at a pressure of 1 atm is 44 L. If the pressure increases to 4 atm, what will its volume be?
Answer:
P₁V₁ = P₂V₂
1 × 44 = 4 × V₂
V₂ = 44/4 = 11L

Question 19.
The volume of a definite mass of gas at 1 atm pressure is 1200 L. How much pressure is applied to change its volume to 30 L? (Temperature constant)
Answer:
P₁ = 1 atm
V₁ = 1200L
V₂ = 30L
P₂ = ?
P₁V₁ = P₂V₂
1 × 1200 = P₂ × 30L
P₂= \(\frac{1200}{30}\) = 40atm
40 atm pressure is needed to change its volume to 30 L

Question 20.
Upto which point of the arrangement does air occupy?
Answer:
It is from the bottom of the bottle to that part of the refill that contains the ink.

Question 21.
What change occurs to the temperature of the air inside the bottle when covering the bottle with your palms?
Answer:
Temperature increases

Question 22.
Does the volume of the gas increase or decrease during this period?
Answer:
Volume increases

Question 23.
Place the bottle in water and cool it. What do you observe?
Answer:
The ink drop moves downward. As the bottle cools, the volume of the air changes.

Question 24.
What happens to the volume of the gas when the bottle is cooled?
Answer:
Volume decreases

Question 25.
From this, can you find out the relation between the temperature of a gas and its volume?
Answer:
As the temperature of the gas increases, volume also increases. When the temperature decreases, volume also decreases.

Towards the end of the 18th century, it was Jacques Alexandre Ceasre Charles, a French scientist, who studied and recorded the changes in the volume of gases in accordance with temperature.

Question 26.
If the volume of a gas is 150 L at 27°C, what will its volume be at 0°C
Answer:
T₁ = 27°C = 27 + 273 = 300K
T₂ = 0°C = 273 + 0°C = 273K
\(\frac{\mathrm{V}_1}{\mathrm{~T}_1}\) = \(\frac{\mathrm{V}_2}{\mathrm{~T}_2}\)
150/300 = V₂/273
V₂ = \(\frac{150 \times 273}{300}\) = 136.5 L

Question 27.
The volume of a definite mass of hydrogen gas at 300 K temperature is 60 mL. At what temperature does the volume of this gas become 20 mL?
Answer:
T₁ = 300K, T₂ = ?
V₁ = 60mL, V₂ = 20mL
\(\frac{\mathrm{V}_1}{\mathrm{~T}_1}\) = \(\frac{\mathrm{V}_2}{\mathrm{~T}_2}\)
\(\frac{600}{300}\) = \(\frac{20}{T_2}\)
T₂ = (20 × 300)/60 = 100K

Question 28.
The relation between volume and temperature of a definite mass of a gas at constant pressure is illustrated in the two graphs given below.
a) What is the relation between temperature and volume?
b) To which gas law do these graphs relate?
c) State the law.
d) What is the peculiarity of the value of V/T?

Answer:
a) As the temperature increases, volume also increases
b) Charles’s law
c) At constant pressure, the volume of a definite mass of a gas is directly proportional to its temperature on the Kelvin Scale. This is Charles’s law.
d) Always Constant

Question 29.
What are the practical applications of Charles’s law in everyday life?
Answer:
During summer, vehicle tyres are filled with air at low pressure.
• Liquid ammonia is a substance that quickly changes from a liquid to a gaseous state. The containers of liquid ammonia are submerged in cold water for some time before opening.

Question 30.
The volume of definite mass of gas at 1 atm pressure and 300K temperature is 30 L. What will be its volume at 273K temperature and 0.5 atm pressure?
Answer:
\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
\(\frac{1 \times 30}{300}\) = \(\frac{0.5 \times V_2}{273}\)
V = \(\frac{30 \times 273}{300 \times 0.5}\) = 54.6 L

Question 31.
The volume of a gas at 27°C and I atm pressure ¡s 100 mL. What will be its volume at 273 K temperature and 2 atm pressure?
Answer:
P₁ = 1 atm, P₂ = 2atm
V₁ = 100mL, V₂ = ?
T₁ = 27°C, T₂ = 273K
\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
\(\frac{1 \times 100}{300}\) = \(\frac{2 \times V_2}{273}\)
V = \(\frac{30 \times 273}{300 \times 0.5}\) = 54.6 L
1 × 100 × 273 = 300 × 2 × V₂
V₂ = \(\frac{100 \times 273}{300 \times 2}\) = 45.5 L

Question 32.
Why is atomic mass a fraction?
Answer:
The atomic mass of elements is calculated by considering the average mass of their isotopes based on their natural abundance.

For example:

• The natural abundance of neon is as, ²⁰Ne = 90.48% , ²¹Ne = 0.27%, Ne = 9.25%
  The average atomic mass of this element = \(\frac{(20 \times 90.48)+(21 \times 0.27)+(22 \times 9.25)}{100}\) = 20.18u
• The natural abundance of Cl-35 isotope is 75% and that of Cl-37 is 25%.
  The average atomic mass ot this element = \(\frac{(35 \times 75)+(37 \times 25)}{100}\) = \(\frac{3550}{100}\) = 35.5u
  The atomic mass of most elements is a fraction since the average atomic mass is calculated in this way.

Question 33.
Why is the quantity of one mole atom of different substances different, even though they contain the same number of atoms?
Answer:
The difference in quantity is due to the different sizes of the atoms, even though they are the same in number.

Question 34.
Calculate the molecular mass of the molecules given in the table. Atomic masses of constituent elements are given in the brackets.
(Atomic mass – H = 1, O = 16, N = 14, C = 12, S = 32)

Answer:

Element /Compound	Chemical formula	Molecular mass
Oxygen	O2	2 × 16 = 32
Ammonia	NH3	14 + 3 × 1 = 17
Water	H2O	1 × 2 + 16 = 18
Glucose	C6H12O6	6 × 12 + 1 × 12 + 16 × 6 = 180
Sulphuric acid	H2SO4	1 × 2 + 32 + 16 × 4 = 98
Nitrogen	N2	14 × 2 = 28

If 6.022 × 10²³ oxygen molecules are taken, the mass will be 32 g. This is the molecular mass of oxygen expressed in grams. This is known as the gram molecular mass.
The amount of a substance in grams equal to its molecular mass is called Gram Molecular Mass (GMM).

Question 35.
What will be the mass of 6.022 × 10²³ CO₂ molecules?
Answer:
44g
This is the molar mass of the compound.
One molar mass of a compound contains one mole of molecules.

Question 36.
Complete the following table.
(Atomic mass-H = 1, O = 16, N = 14, C = 12, S = 32, Ca = 40)

Answer:

Compound	Molecular mass	Molar mass	Number of moles	Number of molecules
NH3	17	17g	1	6.022 × 1023
CO2	44	44g	1	6.022 × 1023
H2O	18	18g	1	6.022 × 1023
NO2	46	46g	1	6.022 × 1023
CaCO3	100	100g	1	6.022 × 1023

Question 37.
How many moles are there in 44g of CO₂?
Answer:
1 mole

Question 38.
How many moles are there in 88g of CO₂
Answer:
88g = \(\frac{88}{44}\) = 2 mole
Number of moles = \(\frac{\text { Given mass }}{\text { Molar mass }}\)

Question 39.
What is the number of molecules in this sample of CO₂?
Answer:
Number of molecules in 1 mole of CO₂ = 6.022 × 10²³
Number of molecules in 2 moles of CO₂ = 2 × 6.022 × 10²³

Question 40.
Complete the following table
(Atomic mass – H = 1 , O = 16, N = 14, C = 12, S = 32, Ca = 40 ,Na = 23 ,C1 = 35.5)

Answer:

Question 41.
What is the peculiarity of the volume occupied by an equal number of molecules of different gases kept at the same temperature and pressure?
Answer:
According to Avogadro’s Law, the peculiarity is that equal numbers of molecules of different gases occupy equal volumes when kept at the same temperature and pressure.

If temperature and pressure are fixed at 273 K and 1 atm respectively, it is known as STP (Standard Temperature and Pressure). At STP, one mole of any gas occupies a volume of 22.4 L. This is known as molar volume at STP.

If temperature and pressure are fixed at 273 K and 1 atm respectively, it is known as STP (Standard Temperature and Pressure). At STP, one mole of any gas occupies a volume of 22.4 L. This is known as
molar volume at STP.

Number of moles of gas at STP = \(\frac{\text { Given volume (in litres) }}{\text { Molar volume at STP }}\)

Question 42.
Complete the table given below

Answer:

Question 43.
224L O₂ (oxygen) gas is taken at STP.
a) How many moles of oxygen are there in it?
b) How many molecules are there in it?
c) Calculate the mass of this oxygen sample. (Hint: atomic mass O = 16)
Answer:
a) No of moles = \(\frac{\text { Volume at STP }}{22.4 \mathrm{~L}}\)
= \(=\frac{224 \mathrm{~L}}{22.4 \mathrm{~L}}\)
= 10 mole

b) No of molecules =No of moles × 6.022 × 10²³ = 10 × 6.022 × 10²³

c) Mass = No of moles × molecular mass
= 10 × 32 = 320g

Question 44.
Find the samples with equal volumes from the ones given below at STP.
a) 64 g O₂
b) 44 g CO₂
c) 2 × 6.022 × 10²³ NH₃ molecules
Answer:
a) 64 g O₂
No of moles = \(\frac{\text { Mass }}{\text { Molecular mass }}\) = \(\frac{64}{32}\) = 2 mole
Volume of Oxygen = No of moles × 22.4L = 2 × 22.4L = 44.8L

b) 44 g CO₂
No of moles \(\frac{\text { Mass }}{\text { Molecular mass }}\) = \(\frac{44}{44}\) = 1 mole
Volume of 1 mole CO₂ = No of moles × 22.4L = 1 × 22.4L = 22.4L

c) 2 × 6.022 × 10²³ NH₃ molecules
No of moles = 2
Volume at STP = No of moles × 22.4L = 2 × 22.4L = 44.8L
Therefore, (a) and (c) have equal volume.

Question 45.
What is the mass of 2 mol of hydrogen?
Answer:
2 × 2 = 4g

Question 46.
What is the mass of 1 mol of oxygen?
Answer:
1 × 32 = 32g
4gH₂ + 32gO₂ → 36gH₂O
According to this equation, when 4 g of hydrogen completely reacts with oxygen, 36 g of water is obtained.

Question 47.
How many grams of water will be obtained if 40 g of hydrogen react completely with oxygen?
Answer:
Water obtained when 4 g H₂ reacts completely = 36g
Water obtained when 1 g H₂ reacts completely = 36/4 = 9g
Water obtained when 40 g H₂ reacts completely = 9 × 40 = 360g

Question 48.
How much oxygen is required for this much hydrogen to react completely and form water?
4gH₂ + 32gO₂ → 36gH₂O
Answer:
The oxygen required for 4 g of hydrogen to react completely = 32g
The oxygen required for 1 g of hydrogen to react completely = 32/4 = 8
The oxygen required for 40 g of hydrogen to react completely = 8 × 40 = 320g

Question 49.
The reaction in which nitrogen and hydrogen react to give ammonia is of high significance in the field of agriculture.
N₂ + 3H₂ → 2NH₃
On the basis of mass, it is 28 g N₂ + 6g H₂ → 34 g NH₃
If we know how much ammonia is produced, we can determine the amount of reactants to be used.
a) How many moles of nitrogen and hydrogen are required to produce 6 mol of ammonia?
1 mol N₂ + 3mol H₂ → 2mol NH₃
Answer:
3 mol N₂ and 9 mol H₂ are required to produce 6 mol NH₃. If the amount of any of these reactants is less, the product will be formed proportionately.

b) What is the ratio of the reactants?
Answer:
1 : 3

c) What is the quantity of ammonia produced when 6 moles of nitrogen react with 6 moles of hydrogen?
Answer:
Only 2 moles of nitrogen react with 6 moles of hydrogen. (N:H = 1:3)
2 moles of nitrogen + 6 moles of hydrogen → 4 moles of ammonia

d) How many moles of nitrogen will be left after the reaction?
Answer:
4 mol

Question 50.
The main component of biogas is methane (CH₄). Note the equation of its combustion.
CH₄ + 2O₂ → CO₂ + 2H₂O
Answer:
On the basis of moles,
1 mol CH₄ + 2mol O₂ → 1 mol CO₂ + 2 mol H₂O
On the basis of mass,
16g CH₄ + 64g O₂ → 44g CO₂ + 36g H₂O

a) What is the mass of oxygen required to react with 16g of methane?
Answer:
64g of Oxygen

b) Find the volume of CO₂ produced by the complete combustion of 16 g of methane.
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O
22.4L CH₄ + 44.8L O₂ → 22.4L CO₂ + 44.8L H₂O

The volume of CO₂ produced by the complete combustion of 16 g of methane is 22.4 L Thus, it can be seen that the amount of CO₂ released when each gram of methane is burned is about three times the mass of methane. Butane (C₄H₁₀) present in cooking gas also emits about three times CO₂ when burned.

Compounds such as lime and baking soda are used to neutralize acid wastes from industries. The above examples will help you to find out the number of compounds to be used for this purpose.

H₂SO₄ + Ca (OH)₂ → CaSO₄ + 2H₂O
H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2H₂O + 2CO₂
The amount of lime and baking soda used by factories can be found out from these equations.

Question 51.
Calculate the amount of lime in kg is required to neutralise 980 kg of sulphuric acid.
Answer:
H₂SO₄ + Ca (OH)₂ → CaSO₄ + 2H₂O
98g H₂SO₄ + 74g Ca (OH)₂ → 136g CaSO₄ – 36g H₂O
(Molecular mass: H₂SO₄ = 98g, Ca (OH)₂ = 74g, CaSO₄ = 136g, H₂O = 36g)
The amount of lime required to neutralise 98g H₂SO₄ is = 74g
Therefore, the amount of lime required to neutralise 980 kg H₂SO₄ is = 740Kg

Std 10 Chemistry Chapter 4 Notes – Extended Activities

Question 1.
448 L HCl gas kept at STP completely reacts with ammonia gas at STP to form NH₄CI.
NH₃ + HCl → NH₄CI
Find the mass of ammonia gas used in this reaction.
Answer:
Moles of HCl = \(\frac{\text { Volume of } \mathrm{HCl}}{\text { Volume at STP }}\) = \(\frac{448 \mathrm{~L}}{22.4 \mathrm{~L}}\) = 20 mole
1 mole of HCl reacts with 1 mole of NH₃.
Moles of NH₃ = Moles of HCl = 20 mol
Molar mass of NH₃ = 14 + (1 × 3) = 17g/mol
Mass of ammonia gas used in this reaction = No of moles of NH₃ × Molar mass of NH₃ = 20 × 17 = 340g

Question 2.
48 g C burns in air, and carbon monoxide gas is formed. If the mass of CO gas formed is 56 g, what is the volume of oxygen gas at STP used in this reaction?
Answer:
2C(s) + O₂(g) → 2CO(g)
Mass of carbon monoxide (CO) formed = 56 g.
Molar mass of carbon monoxide (CO) = 12 + 16 = 28g/mol
Moles of CO = \(\frac{\text { mass of } \mathrm{CO}}{\text { molar mass of } \mathrm{CO}}\) = \(\frac{56 \mathrm{~g}}{28}\) = 2 mol
Moles of O₂ used = 1/2 × moles of CO = 1/2 × 2 mol = 1 mol
Volume of O₂ at STP=No of moles of O₂ molar volume at STP = 1 mol × 22.4mol/L = 22.4L

Gas Laws and Mole Concept Class 10 Notes

Gas Laws and Mole Concept Notes Pdf

• General Properties of Gases :
  • Gases have particles in constant, random motion.
  • Gases are compressible and expand to fill their container.
  • The density of gases is generally lower than liquids and solids.
  • Common gaseous elements at room temperature: Oxygen (O₂), Nitrogen (N₂), Hydrogen (H₂), Helium (He), Neon (Ne), Argon (Ar).
  • Examples of gaseous compounds at room temperature: Carbon dioxide (CO₂), Ammonia (NH₃), Methane (CH₄), Sulphur dioxide (SO₂), Carbon monoxide (CO), Nitrogen dioxide (NO₂).
  • Particle arrangement in gases: large distance between molecules, very low force of attraction, high freedom of movement, and high energy.
• Kinetic Theory of Gases (Postulates):
  • Gases are made of tiny particles.
  • Attractive forces between gas molecules are very low.
  • Volume of gas molecules is negligible compared to the total volume.
  • Volume can be reduced by decreasing the distance between molecules.
  • Gas molecules collide with each other and the container walls, causing pressure.
  • Collisions are elastic (kinetic energy is conserved).
  • Average kinetic energy is directly proportional to temperature.
• Measurable Properties of Gases:
  • Volume (V): Space occupied by the gas (units: L, m³).
  • Pressure (P): Force exerted by gas molecules colliding with the container walls (units: atm, Pa).
  • Temperature (T): Measure of the average kinetic energy of gas molecules (units: K, °C). Remember to convert °C to K by adding 273.
• SI units for volume (m3) and temperature (K).
• Gas Laws
  • Boyle’s Law: At constant temperature and number of moles, the volume of a gas is inversely proportional to its pressure (P₁V₁ = P₂V₂).
  • Charles’s Law: At constant pressure and number of moles, the volume of a gas is directly proportional to its absolute temperature (V₁/T₁ = V₂/T₂). Temperature must be in Kelvin.
  • Avogadro’s Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles (V₁/n₁ = V₂n₂). Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
  • Combined Gas Equation: Relates pressure, volume, and temperature changes for a fixed amount of gas (P₁V₁/T₁C = P₂V₂/T₂).
• Mole Concept
  • Mole: A unit for counting a large number of particles (6.022 × 10²³), known as Avogadro’s number
  • Relative Atomic Mass: The mass of an atom compared to 1/12^th the mass of a carbon-12 atom.
  • Gram Atomic Mass (GAM): The atomic mass of an element expressed in grams. One GAM contains 1 mole of atoms.
  • Molecular Mass: The total mass of atoms in a molecule.
  • Gram Molecular Mass (GMM) / Molar Mass: The molecular mass of a substance expressed in grams per mole (g/mol). One molar mass contains 1 mole of molecules.
• Calculations involving moles:
  • Number of moles = mass / molar mass
  • Number of molecules = moles × Avogadro’s number
  • Number of moles = Volume at STP / 22.4L
• Volume of Gases and Mole :
  • Molar Volume at STP: At Standard Temperature and Pressure (0°C and 1 atm), 1 mole of any gas occupies a volume of 22.4 L.
  • Avogadro’s Law explains this: equal numbers of molecules occupy equal volumes at the same T and P.
• Ideal Gas Equation: Combines Boyle’s, Charles’s, and Avogadro’s laws into the equation: PV = nRT, where:
  • P = pressure
  • V = volume
  • n = number of moles
  • R = Universal Gas Constant (with different values depending on the units of P and V)
  • T = temperature (in Kelvin)
• Ideal Gas: Gases that obey the ideal gas equation at all temperatures and pressures (real gases behave ideally under certain conditions).
• Mole Concept and Chemical Equations :
  • The coefficients in a balanced chemical equation represent the mole ratios of reactants and products.
  • We can use mole ratios to calculate the amounts (in moles, mass, or volume of gases at STP) of substances involved in a reaction.
  • Stoichiometry: The study of the quantitative relationships between reactants and products in chemical reactions.

INTRODUCTION

Numerous phenomena in our daily lives, from weather patterns to the operation of engines, are governed by the principles of gas behaviour. The reason for this behaviour lies in the fundamental properties of gases and how we quantify them. In this unit, we will explore the general properties of gases, the laws that describe their behaviour, and the crucial concept of the mole.

Gases:

• Gases consist of particles that are in constant, random motion.
• Gases can be compressed, meaning their volume can be reduced.
• Gases expand to fill any container they occupy.

Gas Laws:

• Describe the relationships between pressure, volume, and temperature of gases.
• Boyle’s Law explains the inverse relationship between pressure and volume.
• Charles’s Law explains the direct relationship between volume and temperature.
• Avogadro’s law explains the direct relationship between volume and Number of moles

Combined Gas Equation:
• Integrates Boyle’s Law and Charles’s Law to relate pressure, volume, and temperature changes of a gas.

Mole Concept:
• Introduces the “mole,” a unit for counting the number of particles (atoms, molecules) in a substance.

Relative Atomic Mass and Mole:
• Relative atomic mass is the mass of an atom compared to the mass of a carbon-12 atom

Molar Mass:
• Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

Volume of Gases and Mole:
• The molar volume of a gas at STP allows us to relate the number of moles of a gas to its volume.

Ideal Gas Equation:
Relates pressure, volume, temperature, and the number of moles of a gas in the equation PV = nRT.

Mole Concept and Chemical Equations:

• Applies the mole concept to calculate the amounts of gases involved in chemical reactions.
• Allows us to predict the volumes of gases consumed or produced in a reaction.

GENERAL PROPERTIES OF GASES
As in this picture, we can see balloons flying in the air in some tourist centres.

MAIN POSTULATES OF THE KINETIC THEORY OF GASES

• Gases are made up of minute particles (atoms/molecules).
• The attractive force between molecules is very low.
• As the molecules are so far apart, the volume of gaseous molecules is negligible in comparison with the total volume of the gas.
• The volume can be reduced by reducing the distance between the gaseous molecules.
• Gaseous molecules collide with one another and with the walls of the container due to their constant motion in all directions. The force produced due to the collision of molecules with the walls of the container results in gaseous pressure.
• Collisions of gaseous molecules are elastic in nature. That means the kinetic energy of the molecules before and after the collision will be the same.
• The average kinetic energy of the gaseous molecules is directly proportional to their temperature.

GENERAL PROPERTIES OF GAS
The measurable properties of gas, such as volume, pressure and temperature, are explained below.

VOLUME
The space occupied by a substance is taken as its volume.

Units of volume
Generally, the unit used is litre (L).
The SI unit of volume is m³

1000cm3 = 1000mL = 1L

• 1cm³ is the volume of a container having 1cm length, breadth 1cm and height 1cm.
• 1m³ is the volume of a container having lm length, breadth 1m and height 1m.
  1m³ = 1m × 1m × 1m
  = 100cm × 100cm × 100cm = 1000000cm³
  = 1000000mL
  \(\frac{1000000}{1000}\) = [1000mL = 1L]
  1000L
  ∴ 1m³ = 1000L

PRESSURE
Gaseous molecules are in constant random motion. As a result, they collide with one another and with the walls of the container. Due to these collisions, a force is experienced on the surface.

Gaseous pressure is the force experienced per unit surface area. Pressure = \(\frac{\text { Force }}{\text { Surface area }}[latex] = [latex]\frac{\dot{F}}{A}[latex]

Units of pressure
Generally, pressure is expressed in terms of atmospheric pressure (atm)
The SI unit of pressure is Pascal (Pa)
(Pa = N/m or Newton per square meter)
1 atm = 1.01325 × 10⁵ pa

Pressure of gases is measured using manometers.

TEMPERATURE
Temperature is another measurable property of gas.

Units of temperature
The SI unit of temperature is Kelvin(K).
• To convert the common unit °C to Kelvin, add 273 to it.
If the temperature is t°C,

t°C = (t + 273) K

GAS LAWS
BOYLE’S LAW – The relation between pressure and volume of a gas
Take a large syringe and remove its piston. Put into the syringe a small, inflated balloon with its open end tied up. Now, refix the piston in its position. Close the other end of the syringe with your finger. Record the changes happening to the balloon when the piston is pushed in and pulled back.

Activity	Observation
The piston is pushed in.	Balloon shrinks
The piston is pulled back.	Balloon expands

CHARLES’S LAW – The relation between volume and temperature
Let’s do an experiment. Make a small hole in the lid of a glass bottle and fix an empty refill into it. Add a small drop of ink into the refill, then blow gently to move the ink to the centre of the refill. Cover the bottle with your palms. What do you observe?
Answer:
The ink drop in the refill rises to the top.

Volume – temperature graph
Plot the relation between the temperature and volume of gases at constant pressure in a graph. If this graph is extrapolated backwards, it will meet the temperature axis (X axis) at – 273.15 °C.

This means that at -273.15°C, the volume of a gas becomes zero.
Even if the pressure is changed, it is found that the temperature-volume graph is a straight line, and all the straight lines meet at – 273.15 °C.

The analysis of the graph is given in the table below.

	Kelvin scale	Celsius scale	Volume
Absolute zero	0K	-273°C	0
Freezing point of water	273K	0°c	115L
Boiling point of water	373K	100°C	150L

According to the table, it is evident that as temperature increases, the volume of gas increases.
Absolute zero Temperature:
It was Lord Kelvin who identified that the lowest temperature that can be attained by a gas is – 273.15 °C and named it Absolute zero. For practical purposes, – 273 °C is taken as the value of Absolute zero. Using this, he developed the Kelvin scale of temperature.

At constant pressure, the volume of a definite mass of a gas is directly proportional to its temperature on the Kelvin Scale. This is Charles’s law. V ∝ T (pressure, mass constant) V = k xT (k – a constant) V/T = k, a constant If the volumes of a definite mass of gas are V1 and V2 at temperatures T1 and T2, respectively, then. V1/T1 = V2/T2

AVOGADRO’S LAW -The relation between the number of particles in gases (N) and their volume (V).

At constant temperature and pressure, equal volumes of all gases contain an equal number of molecules. In other words, at constant temperature and pressure, equal numbers of molecules of different gases occupy equal volumes.

At constant temperature and pressure, the volume of a gas is directly proportional to the number of molecules. This is Avogadro’s Law. V ∝ N (Temperature, pressure constant)

Applications of Avogadro’s law in daily life.

• Inflating balloons.
• Filling air in footballs.

COMBINED GAS EQUATION
Boyle’s law
V ∝ 1/p (temperature, mass constant)

Charles’s law
V ∝ T (pressure, mass constant) Considering both laws together,
V ∝ 1/p × T
V = a constant × 1/p × T
[latex]\frac{\mathrm{PV}}{\mathrm{~T}}\) = k (a Constant)
If the pressure, volume and temperature of a definite mass of a gas are changed from P₁, V₁), and T₁, to P₂, V₂ and T₂ respectively, then.
\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
This is a combined gas equation.

MOLE CONCEPT
In everyday life, various units are used to count objects:
Eg.
2 numbers: pair 12 numbers: dozen
20 numbers: score 144 numbers: gross

Are these units sufficient to count extremely small particles like atoms, molecules and ions?
Answer:
No
Even the presence of such tiny particles can be detected only with the help of powerful modem microscopes. Counting such minute particles is impossible.
Mole is the unit used to indicate such large numbers.

MOLE:-
A mole is a quantity of a substance containing 6.022 × 10²³ particles (atoms/molecules/ions). This number came to be known as the Avogadro number (N_A), named in honour of the Italian scientist Amedeo Avogadro.

Mole is the SI unit of quantity of matter. This represents the number of particles in that substance.
One mole of water contains 6.022 × 10²³ water molecules.
The concept of the mole is highly significant in Chemistry. It enables accurate measurement of the quantity of reactants and products that are to be used in chemical reactions.

1 Mole = 6.022 × 1023 1NA = 6.022 × 1023

RELATIVE ATOMIC MASS AND MOLE

• Relative atomic mass expresses the mass of one atom relative to that of another.
• It indicates how many times one atom is heavier as compared to another. The atomic mass of an element is expressed as how many times the mass of the atom is when compared to the 1/12th mass of a carbon- 12 atom, which is considered a single unit. This mass is known as the unified mass.
• The atomic masses have decimal values because the average atomic mass is calculated by taking into account the presence of isotopes. However, for practical purposes, they are often used as whole numbers.

Element	Average atomic mass	Relative atomic mass
Hydrogen	1.0079	1
Oxygen	15.9994	16
Sodium	22.989	23
Carbon	12.011	12
Nitrogen	14.0067	14

Gram Atomic Mass
Analyse the table given below.

Element	Relative atomic mass	Gram atomic mass	Number of mole atoms	Number of atoms
Copper	63.5	63.5g	1	6.022 × 1023
Iron	55.8	55.8g	1	6.022 × 1023
Zinc	65.3	65.3g	1	6.022 × 1023
Aluminum	27	27g	1	6.022 × 1023
Nitrogen	14	14g	1	6.022 × 1023

An element that weighs as much as its relative atomic mass in grams contains 6.022 × 10²³ atoms.
This mass is known as the gram atomic mass (GAM). One-gram atomic mass contains 1 mole of atoms.

MOLAR MASS
The total mass of atoms in a molecule is known as molecular mass.
Eg. Calculate the molecular mass of carbon dioxide (CO₂, atomic mass O = 16, C = 12).
Molecular mass = 1 × C + 2 × O
= 1 × 12 + 2 × 16 = 12 + 32 = 44

VOLUME OF GASES AND MOLE
According to Avogadro’s law, at constant temperature and pressure, equal volumes of all gases contain equal numbers of molecules.

IDEAL GAS EQUATION
Considering Boyle’s law, Charles’s law and Avogadro’s law,
V ∝ 1/P (T, n constant)
V ∝ T (P, n constant
V ∝ n (n=number of moles), (P, T constant)
V ∝ 1/P × T × n
PV = a constant × nT
This constant is known as the Universal Gas Constant. This is represented by the letter R.
PV = n R T
This is an ideal gas equation.
The gases that obey the ideal gas equation at all temperatures and pressures are known as ideal gases.

MOLE CONCEPT AND CHEMICAL EQUATION
Here, it can be seen that two hydrogen molecules combine with one oxygen molecule to form two water molecules.
2H₂ + O₂ → 2H₂O
If we take two moles of hydrogen molecules instead of two hydrogen molecules
2 mol H₂ + 1 mol O₂ → 2 mol H₂O