Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m−2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
Q. Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m−2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

(A) 0.5217

(B) 0.480

(C) 0.340

(D) 0.663

Correct Answer: 0.5217

Explanation

For an ideal solution, Raoult’s law applies.

Partial pressures:

PA = xA · PA° PB = xB · PB°

Total vapour pressure:

P = PA + PB

Given mole fraction of A in vapour:

yA = PA / P = 0.8

So,

0.8 = (xA · 55) / [xA · 55 + (1 − xA) · 15]

Solving,

0.8(55xA + 15 − 15xA) = 55xA 0.8(40xA + 15) = 55xA 32xA + 12 = 55xA 23xA = 12 xA = 12 / 23 = 0.5217

Hence, the mole fraction of A in the solution is 0.5217.

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