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Two liquids A and B have vapour pressures in the ratio $P_{A}^{O} : P_{B}^{O} = 1 : 3$ at certain temperature. Assume A & B form ideal solution and the ratio of mole fractions of A and B in the vapour phase is 4:3, then the mole fraction of ‘B’ in the solution at the same temperature is

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$\frac{1}{5}$

$\frac{2}{3}$

$\frac{4}{5}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$P_{A}^{1} = \frac{P_{A}}{P_{A} + P_{B}}, P_{B}^{1} = \frac{P_{B}}{P_{A} + P_{B}}, P_{A} = x_{A} . P_{A}^{0}, P_{B} = x_{B} . P_{B}^{0}$

$\frac{P_{A}^{1}}{P_{B}^{1}} = \frac{X_{A} . P_{A}^{0}}{X_{B} . P_{B}^{0}} \Rightarrow \frac{4}{3} = \frac{x_{A} \times 1}{x_{B} \times 3} \Rightarrow \frac{x_{A}}{x_{B}} = \frac{4}{1}$

$\frac{X_{A}}{1 - X_{A}} = 4 \Rightarrow 5 X_{A} = 4 \Rightarrow X_{A} = \frac{4}{5}$ and $x_{B} = \frac{1}{5}$

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