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Q.

Two liquids A and B have $P_{A}^{\circ}$ and $P_{B}^{\circ}$ in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phase are 1 : 3 then mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to

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a

0.1

b

1.0

c

0.2

d

0.5

answer is A.

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Detailed Solution

$\begin{array}{l} x_{A} = \frac{1}{4}, x_{B} = \frac{3}{4} y_{A} = \frac{p_{A}^{\circ} x_{A}}{p_{T}} - (1) y_{B} = \frac{p_{B}^{\circ} x_{B}}{p_{T}} - (2) \frac{y_{A}}{y_{B}} = \frac{p_{A}^{\circ}}{p_{B}^{\circ}} \frac{x_{A}}{x_{B}} \frac{y_{A}}{1 - y_{A}} = \frac{1}{3} \times \frac{1}{3} 9 y_{A} = 1 - y_{A} \\ y_{A} = 0.1 \end{array}$

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Two liquids A and B have PA∘ and PB∘ in the ratio of 1 : 3 and the ratio of number of moles of A and B in liquid phase are 1 : 3 then mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to