Question

Two liquids A and B have $P_A^o$ and $P_B^o$ in the ratio $1:3$ 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:

• A. $0.1$
• B. $0.2$
• C. $0.5$
• D. $1.0$

Answer 1

The correct option is A $0.1$

Partial pressure of $A=K\times \frac{1}{4},$ where $K$ is ratio constant.)

Partial pressure of $B=3K\times \frac{3}{4}$

Total pressure is $\frac{10K}{4}$.

Vapor phase mole fraction of $A$ is,

$\frac{\text{Partial pressure of}A}{\text{Total pressure}}=\frac{K\times \frac{1}{4}}{\frac{10K}{4}}=\frac{1}{10}=0.1$