Liquids A and B form and ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are $7 \times 10^{3} \mathrm{~Pa}$ and $12 \times 10^{3} \mathrm{~Pa}$, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of $A$ at this temperature is

(B) $\mathrm{x}_{\mathrm{A}}=0.28 ; \mathrm{x}_{\mathrm{B}}=0.72$

(A) $\mathrm{X}_{\mathrm{A}}=0.37 ; \mathrm{X}_{\mathrm{B}}=0.63$

(C) $\mathrm{X}_{\mathrm{A}}=0.4 ; \mathrm{X}_{\mathrm{B}}=0.6$

(D) $\mathrm{x}_{\mathrm{A}}=0.76 ; \mathrm{x}_{\mathrm{B}}=0.24$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Solutions

Liquids A and B form and ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are

7 \times 1 0^{3} Pa

and

12 \times 1 0^{3} Pa

, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of

A

at this temperature is

Answer: (B)

For an ideal solution each partial pressure follows Raoult's law,

P_{i} = x_{i} P_{i}^{\circ}

. The liquid is

40%

A, so the total pressure is

0.4 \times 7 \times 1 0^{3} + 0.6 \times 12 \times 1 0^{3} = 2.8 \times 1 0^{3} + 7.2 \times 1 0^{3} = 1 0^{4}

Pa. The vapour mole fraction of A is its partial pressure over the total,

\frac{2.8 \times 1 0 ^{3}}{1 0 ^{4}} = 0.28

, leaving B at

0.72

. The vapour is richer in B than the liquid because B is the more volatile component here.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option