NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT TEXTBOOK QUESTIONS SOLVED ( States of Matter )

Question 1. What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?
Answer: P₁ = 1 bar,P₂ = ?       V₁= 500 dm³ ,V₂=200 dm³
As temperature remains constant at 30°C,
P₁V₁=P₂V₂
1 bar x 500 dm³ = P₂ x 200 dm³ or P₂=500/200 bar=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 mL at 35°C. What would be its pressure?
Answer:  V₁= 120 mL, P₁=1.2 bar,
V₂ = 180 mL, P₂ = ?
As temperature remains constant, P₁V₁ = P₂V₂
(1.2 bar) (120 mL) = P2 (180mL)

Question 4.  At 0°C, the density of a gaseous oxide at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer: Using the expression, d =MP/RT , at the same temperature and for same density,
M₁P₁ = M₂P₂ (as R is constant)
(Gaseous oxide) (N₂)
or
M₁ x 2 = 28 x 5(Molecular mass of N₂ = 28 u)
or M₁ = 70u

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 0.15g of aluminium reacts?
Answer: The chemical equation for the reaction is
2 Al + 2 NaOH + H₂0 -> 2 NaAl0₂ + 3H₂ (3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H₂ gas = 3 x 22400 0.15 g of Al at N.T.P release

Question 8. What will be the pressure of the gas mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?
Answer: Calculation of partial pressure of H₂ in 1L vessel P₁= 0.8 bar,
P₂= ? V₁= 0.5 L , V₂ = 1.0 L
As temperature remains constant, P₁V₁ = P₂V₂
(0.8 bar) (0.5 L) = P₂ (1.0 L) or P₂ = 0.40 bar, i.e., PH₂ = 0.40 bar
Calculation of partial pressure of 02 in 1 L vessel
P₁‘ V₁ = P₂‘V₂‘
(0.7 bar) (2.0 L) = P₂ (1L) or P₂‘ = 1.4 bar, i.e.,Po₂= 1.4 bar
Total pressure =P_Hz + P_Q2 = 0.4 bar + 1.4 bar = 1.8 bar

Question 21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Answer: At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

Question 22. Critical temperature for CO₂ and CH₄ are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer:  Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, Co₂ has stronger intermolecular forces than CH₄.

Question 23. Explain the physical significance of vander Waals parameters.
Answer: ‘a’ is a pleasure of the magnitude of the intermolecular forces of attraction, while b is a measure of the effective size of the gas molecules.