Question

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^{\circ }:P_B^{\circ }=1:2$ at a certain temperature. Suppose that we have an ideal solution of $A$ and $B$ in the mole fraction ratio $A:B=1:2,$ the mole fraction of $A$ in the vapour in equilibrium with the solution at the given temperature is

• A. 0.25
• B. 0.2
• C. 0.5
• D. 0.33

Answer 1

Answer: (B)

Since the ratio of $P_A^{\circ }$ and $P_B^{\circ }$ is 1: 2 and mole fraction is 1: 2 therefore,

partial pressure of $A\left(P_A^{\prime }\right)=P_A^{\circ }{x}_A$ and

partial pressure of $B\left(P_B^{\prime }\right)=P_B^{\circ }{x}_B$ are related as

$P_B^{\prime }=4P_A^{\prime }$

$P=P_A^{\prime }+P_B^{\prime }=P_A^{\prime }+4P_A^{\prime }=5P_A^{\prime }$

The mole fraction of $A$ in the vapour in equilibrium with solution (according to Dalton's law of partial pressure) is

$x_A^{\prime }=\frac{P_A^{\prime }}{P}=\frac{P_A^{\prime }}{5P_A^{\prime }}=\frac{1}{5}=0.2$