NIOS Class 12 Chemistry Chapter 5 The Gaseous and Liquid State

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NIOS Class 12 Chemistry Chapter 5 The Gaseous and Liquid State

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The Gaseous and Liquid State

Chapter: 5

Module – III: States of Matter

INTEXT QUESTION 5.1

1. The density of a gas is usually less than that of the liquid. Explain.

Ans: Due to more intermolecular distances in gaseous molecule compared to liquid.

2. Calculate the pressure (atm) required to compress 500 mL of gas at 0.20 atm

into a volume of 10 mL.

Ans: Boyle Law equation is:

p₁V₁ = p₂V₂

(.20 atm) (500 mL) = p₂ (10 mL)

p₂ = (0.20 atm) (500 ml)/(10 ml)

p₂ = 10 atm.

3. Equal volumes of oxygen gas and an unknown gas weigh 2.00 and 1.75 grespectively under the same experimental conditions. What is the molar mass of the unknown gas?

Ans: By Avogadro’s Law:

moles of O₃ = moles of unknown gas

(2.00 g)/(32 g mole⁻¹) = (1.75 g)/(Molecular weight of unknown gas)

Molar mass of unknown gas = (1.75 ×32)/(2.00) = 28 g mol⁻¹

Molar mass of unknown gas is 28.

4. What type of intermolecular interactions are present in (a) Ne gas (b) Carbon monoxide?

Ans: (a) Dispersion or London Forces (b) dipole-dipole interactions and dispersion forces.

INTEXT QUESTIONS 5.2

1. What is the difference between diffusion and effusion.

Ans: Movement of gas molecules through another gas is called diffusion. When gas escapes from a container through a very small opening it is called effusion.

2. Explain why Daltons’ law is not applicable to a system of ammonia and

hydrogen chloride gas.

Ans: Ammonia and hydrogen chloride gases are reacting gases and Dalton’s Law is applicable to mixture of non-reacting gases.

3. The rates of diffusion of CO₂ and O₃ were found to be 0.29 and 0.271. What is the molecular mass of O₃ if the molecular mass of CO₂ is 44.

Ans: (Rₒ₃)/{r(co₂)} = [{ M(co₃)}\(Mₒ₃)]¹/²

(0.271)/(0.290) = [(44)/(Mₒ₃)]¹/²

Squaring both sides

(0.271)²/(0.290)² = (44)/(Mₒ₃)

Mₒ₃ = (44× 0.29× 0.29)/(0.271×0.271)

= 50.4

Molecular mass of O₃ = 50.4

4. Calculate the pressure exerted by 5.0 mol of carbon dioxide in a 1 litre flask

at 47ºC using ideal gas equation.

Ans: By ideal gas equation:

pV = nRT

p × 1.0 = (5.0 mol) (0.0821 L atm K⁻¹mol⁻¹) 320 K

p = (5.0 mol) (0.0821 L atm K⁻¹ mol⁻¹ ) (320 K)/(1.0 L)

p = 131.3 atm

INTEXT QUESTIONS 5.3

1. What are the conditions under which real gases behave as ideal gas.

Ans: Real gases behave like ideal gases under conditions of low pressure and high temperature because:

(i) At low pressure, the molecules are far apart, so intermolecular forces become negligible.

(ii) At high temperature, the kinetic energy of gas molecules is very high, which overcomes any intermolecular attraction.

2. Which term in van der waals equation accounts for the molecular volume.

Ans: b.

3. Calculate the root mean square velocity of ozone kept in a closed vessel at 20º C and 1 atm pressure.

Ans: Uᵣₘₛ = √(3RT)/(M)

= √3 (8.314 Jk⁻¹ mol⁻¹ ) (293 K)/(0.048 kg mol⁻¹ )

= √(8.314 Kg m² s⁻² K⁻¹ mol⁻¹ ) (293K)/(0.048 Kg mol⁻¹)

= 390.3 ms^–1.

4. What is compressibility factor.

Ans: Z = (pVm) /(RT) = where Vm = molar volume

Z is compressibility factor.