Question

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^{\circ }:P_B^{\circ }=1:3$ at a certain temperature. Assume $A$ and $B$ form an ideal solution and the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4:3$. Then the mole fraction of $B$ in the solution at the same temperature is:

• A. $\frac{1}{5}$
• B. $\frac{2}{3}$
• C. $\frac{4}{5}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{5}$

By using Raoult's law:
$y_A=\frac{P_A}{P}=\frac{P_A^{\circ }{x}_A}{P}$ and $y_B=\frac{P_B^{\circ }{x}_B}{P}$
Where,
$P_A$ is vapour pressure of liquid $A$
$P_B$ is vapour pressure of liquid $B$
$P_A^{\circ }$ is vapour pressure of pure liquid $A$
​​​​​​​$P_B^{\circ }$ is vapour pressure of pure liquid $B$
$P$ is the total vapour pressure of the liquid
$y_A$ and $y_B$ are the mole fractions of $A$ and $B$ in vapour phase. $x_A$ and $x_B$ are mole fractions of $A$ and $B$ in solution.
We know that $x_A+x_B=1$
Dividing $y_A$ by $y_B$, we get
$\frac{y_A}{y_B}=\frac{P_A^{\circ }}{P_B^{\circ }}\times \frac{x_A}{x_B}\Rightarrow \frac{4}{3}=\frac{1}{3}\times \frac{x_A}{(1-x_A)}$
$\Rightarrow x_A=\frac{4}{5}$
$\therefore x_B=\frac{1}{5}$