Gujarat Board GSEB Textbook Solutions Class 11 Chemistry Chapter 5 States of Matter Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Chemistry Chapter 5 States of Matter

### GSEB Class 11 Chemistry States of Matter Text Book Questions and Answers

Question 1.

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 to 30°C ?

Solution

P_{1} = 1 bar, P_{2} = ?

V_{1} = 500 dm , V_{2} = 200 dm³

Temperature remains constant at 30°C

∴ According to Boyle’s Law

1 x 500 = P_{2} x 200

or P_{2} = \(\frac { 1×500 }{ 200 }\) = 2.5 bar.

Question 2.

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 L at 35°C. What would be its pressure ?

Solution:

Here P_{1} = 1.2 bar, P_{2} = ?,

V_{1} = 120 mL, V_{2} = 180 mL

Temperature remains constant at 35°C

∴ According to Boyle’s Law

P_{1} V_{1} = P_{2} V_{2}

1.2 X 120 = P_{2} x 180

or P_{2} = \(\frac { 1.2×120 }{ 180 }\) = 0.8 bar

Question 3.

Using the equation of state pV = nRT; show that a given temperature density of a gas is proportional to gas pressure p.

Solution:

Hence density (ρ) of a gas ∝ p [ ∵ RT m are constants]

Question 4.

At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?

Solution:

Density of nitrogen, ρ = \(\frac { Pm }{ RT }\)

where, p = Pressure, m = Molecular mass, R = Gas constant, T = Temperature

Now, P = 5 x 0.987 atm [∵ 1 bar = 0.987 atm]

m = 28, R = 0.0821 L atm/K/mol.

∴ ρ = \(\frac{5 \times 0.987 \times 28}{0.0821 \times 273}\) [∵ T = 0°C + 273 = 273 K]

Because density of the gaseous oxide is the same

∴ \(\frac{5 \times 0.987 \times 28}{0.0821 \times 273}\) = \(\frac{2 \times 0.987 \times x}{0.0821 \times 273}\)

∴ x = \(\frac { 5×28 }{ 2 }\) = 70 g mol^{-1}

∴ Molecular mass of the oxide = 70.0 g mol^{-1}

Question 5.

Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Solution:

Pressure of 1 gm of an ideal gas A at 27°C (300 K) = 2 bar Pressure on introducing 2 gm of another ideal gas ‘B’ in the same flask = 3 bar

P_{A} = 2 bar; P_{B} = 3 – 2 = 1 bar

Let molecular masses of A and B be M_{A} and B_{B} respectively.

Question 6.

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?

Solution:

The reaction is

Volume of H_{2} released = 202.5 mL.

Question 7.

What will be the pressure exerted by a mixture of 3.2 g of methane and.4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C ?

Solution:

Molar mass of methane (CH_{4}) = 12 + 4×1 = 16 g mol^{-1}

Mass of carbon dioxide (CO_{2}) = 12 + 2 x 16 = 44 g mol^{-1}

Total pressure of mixture

= pCH_{4} + pCO_{2} = (5.54 + 2.77) 10^{4} Pa

= 8.32 x 10^{4} Pa.

Question 8.

What will be the pressure of the gaseous mixture when 0.5 L of H_{2} at 0.8 bar and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel at 27°C ?

Solution:

To calculate the partial pressure of H_{2}, i.e., pH_{2}

V_{1} = 0.5 L V_{2} = 1L

P_{1} = 0.8 bar P_{2} = ?

Temperature remaining constant, applying Boyle’s Law

P_{1} V_{1} = p_{2} V_{2}

0.8 x 0.5 = P_{2} x 1 or P_{2} = 0.4 bar

To calculate the partial pressure of O_{2} i.e., pO_{2}

V_{1} = 2.0 L V_{2} = 1 L

P_{1} = 0.7 bar P_{2} = ?

Applying Boyle’s Law

p_{1}V_{1} = P_{2}V_{2}

0.7 x 2.0 = 1 x P_{2}

or P_{2} = 1.4 bar

If P_{1} is final pressure of the gas mixture, then according to Dalton’s Law of partial pressures

P = pH_{2} + pO_{2} = (0.4 + 1.40) bar = 1.80 bar.

Question 9.

Density of a gas is found to be 5.46 g/dm³ at 27°C at 2 bar pressure. What will be its density at STP?

Solution:

Density of the gas at STP = 3 g/dm³.

Question 10.

34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?

Solution:

P = 0.1 bar

V = 34.05 mL = 0.03405 dm³

R = 0.083 bar dm³K^{-1} mol^{-1}

T = 546 + 273 = 819 K

Now PV = nRT

or n = \(\frac { PV }{ RT }\) = \(\frac{0.1 \times 0.03405}{0.083 \times 819}\) = 0.00005 mol^{-1}

∴ Molar mass of phosphorus = \(\frac { 0.0625 }{ 0.00005 }\) = 1250 g mol^{-1}.

Question 11.

A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out ?

Solution:

T_{1} = 27 + 273 = 300 K, T_{2} = 477 + 273 = 750 K

Applying Charles Law :

\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) [As volume remains same]

or \(\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}\) = \(\frac { 300 }{ 750 }\) = \(\frac { 2 }{ 5 }\)

Thus fraction of air expelled = 1 – \(\frac { 2 }{ 5 }\) = \(\frac { 3 }{ 5 }\) = 0.6.

Question 12.

Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.

Solution:

P = 3.32 bar

V = 5 dm³, R = 0.083 bar dm³ K^{-1}mol^{-1}

n = 4 moles

Since

PV = nRT

T = \(\frac { PV }{ nR }\)

= \(\frac{3.32 \times 5}{4 \times 0.083}\)

= \(\frac { 16.6 }{ 0.332 }\)

= 50 K.

Question 13.

Calculate the total number of electrons present in 1.4 g of nitrogen gas.

Solution:

28 g of N_{2} gas = 1 mole

1.4 g of N_{2} gas = \(\frac { 1.4 }{ 28 }\) = \(\frac { 1 }{ 20 }\) mole

Total no. of electrons present in \(\frac { 1 }{ 20 }\) or 0.05 mol of N_{2}

= 0.5 x 6.02 x 10^{23} x 14 electrons

= 4.2 x 10^{23} electrons [ ∵ N_{2} molecule contains 14 electrons]

Question 14.

How much time would it take to distribute one Avogadro number of wheat grains, in 1010 grains are distributed each second?

Solution:

10^{10} grains are distributed in 1 second

∴ 6.02 x 10^{23} grains are distributed in

Question 15.

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C R = 0.0833 bar dm³ K^{-1}mol^{-1}.

Solution:

Wt. of O_{2} = 8.0 gm; Moles of O_{2} = \(\frac { 8 }{ 32 }\) = \(\frac { 1 }{ 4 }\)

Wt. of H_{2} = 4.0 gm; Moles of H_{2} = \(\frac { 4 }{ 2 }\) = 2

∴ Total moles present in the mixture = \(\frac { 1 }{ 4 }\) + 2 = \(\frac { 9 }{ 4}\)

Volume of the vessel = 1 dm³

Now PV = \(\frac { nRT }{ V }\) = \(\frac { 9 }{ 4 }\) x \(\frac { 0.083×300 }{ 1 }\) = 56.025 bar.

Question 16.

Pay load is defned asthe difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass loo kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^{-3} and R = 0.083 bar dm³ K^{-1} mol^{-1}).

Solution:

Volume of balloon, V = \(\frac { 4 }{ 3 }\)πr³

Radius of the balloon = 10 m

∴Volume of the balloon = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) (10 m)³

= 4190.5 m³

Volume of the filled at 1.66 bar and 27°C = 4190.5 m³

Calculation of mass of He

Total mass of balloon loaded with He

= 100 + 1117.5 = 1217.5 kg

Maximum mass of the air that can be displaced by balloon to go = Volume x density

= 4190.5 m³ x 1.2 kg m^{-3} = 5028.6 kg

∴Payload = (5028.6 – 1217.5) kg = 3811.1 kg.

Question 17.

Calculate the volume occupied by 8.8 g of CO_{2} at 31.1°C and 1 bar pressure R = 0.083 bar LK^{-1} mol^{-1}.

Solution:

P = 1 bar, T = 31.1 + 273 = 304.1 K,

\(\mathrm{W}_{\mathrm{CO}_{2}}\) = 8.8 g, \(\mathrm{M}_{\mathrm{CO}_{2}}\) = 44

∴ PV = \(\frac { W }{ m }\)RT

1 x V = \(\frac { 8.8 }{ 44 }\) x 0.083 x 304.1

or V = 5.05 L.

Question 18.

2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molarmass of the gas?

Solution:

Applying the ideal gas equation PV = nRT

Case I. Let the molar mass of air = M g mol^{-1}

Mass of gas = 2.9 g

No. of moles m = \(\frac{\text { Mass }}{\text { Molar mass }}\) = \(\frac { 2.9 }{ M }\) mol

T = 273 + 95 = 368 K

PV = \(\frac { 2.9 }{ M }\) x R x 368 … (1)

Case II. Mass of hydrogen = 0.184 g

No. of Moles of H_{2} = \(\frac { 0.184 }{ 2 }\)

T = 273 + 17 = 290 K

PV = \(\frac { 0.184 }{ 2 }\) x R x 290 … (2)

From (1) and (2)

\(\frac { 2.9 }{ M }\) x 368 = \(\frac { 0.184 }{ 2 }\) x 290

M = \(\frac{2.9 \times 368 \times 2}{0.184 \times 290}\) = 400 g mol^{-1}

Question 19.

A mixture of dihydrogen and dioxygen at . one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Solution:

If mass of H_{2} = 20 g, than O_{2} = 80 g

Question 20.

What would be the SI unit for the quantity pV²T²/n?

Solution:

Question 21.

In terms of Charles’ law explain why -273°c is the lowest possible temperature.

Solution:

According to Charle’s Law

Thus, – 273° C is the lowest temperature for a gas because below this temperature, its volume will become negative. This is meaningless.

Question 22.

Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?

Solution:

Higher the critical temperature more easily the gas can be liquified, that is, greater are the intermolecular forces of attraction. Hence CO_{2} has stronger intermolecular forces than CH_{4}.

Question 23.

Explain the physical significance of van der Waals parameters.

Solution:

Significance of constant ‘b’. The constant ‘V is called co-volume or excluded volume per mol of a gas. Its units are litre mol^{-1}. The volume of ‘b’ is four times the actual Volume of the molecules It is measure of effective size of the gas molecules.

Significance of constant ‘a’. The value of constant ‘a’ is a measure of the magnitude of intermolecular forces between the molecules of the gas. Its units are atm L² mol^{-2}. Larger the value of V larger will be the intermolecular forces among the gas molecules.