Two liquids A and B form an ideal solution. At 350 K, the vapour pressures of pure A and pure B are 7times10^3 Pa and 12times10^3 Pa respectively. For a solution with 40 mol% A, what is the composition of the vapour in equilibrium?

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EEJ MAIN Chemistry QUESTION #7528

Question 1

Two liquids A and B form an ideal solution. At 350 K, the vapour pressures of pure A and pure B are $7\times10^3$ Pa and $12\times10^3$ Pa respectively. For a solution with 40 mol% A, what is the composition of the vapour in equilibrium?

$x_A=0.37;\ x_B=0.63$
$x_A=0.28;\ x_B=0.72$✔️
$x_A=0.4;\ x_B=0.6$
$x_A=0.76;\ x_B=0.24$

Correct Answer Explanation

Mole fraction in liquid: $x_A = 0.40,\ x_B = 0.60$

By Raoult's Law:

$P_A = x_A \cdot P_A^0 = 0.40 \times 7000 = 2800\ \text{Pa}$

$P_B = x_B \cdot P_B^0 = 0.60 \times 12000 = 7200\ \text{Pa}$

$P_{\text{total}} = 2800 + 7200 = 10000\ \text{Pa}$

In vapour phase: $y_A = \dfrac{P_A}{P_{\text{total}}} = \dfrac{2800}{10000} = \mathbf{0.28}$

$y_B = 1 - 0.28 = \mathbf{0.72}$

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