Question

Two liquids A and B have vapour pressure in the ratio $P_A^{\circ }:P_B^{\circ }=1:3$ at a cetrain temperature. Assume A and B from an ideal solution and the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4:3$. Choose the correct statement:

• A. The mole fraction of $B$ in the solution at the same temperature is $\frac{1}{5}$
• B. The mole fraction of $A$ in the solution at the same temperature is $\frac{4}{5}$
• C. The mole fraction of $A$ in the solution at the same temperature is $\frac{1}{5}$
• D. The mole fraction of $B$ in the solution at the same temperature is $\frac{4}{5}$

Answer 1

The correct option is A The mole fraction of $B$ in the solution at the same temperature is $\frac{1}{5}$

The mole fraction of A in vapour phase is
$y_A=\frac{P_A}{P}=\frac{P_A^{\circ }{x}_A}{P}$
Where, $P_A=\text{Partial pressure of vapour A}$
$x_A=\text{Mole fraction of A}$
The mole fraction of B in vapour phase is
$y_B=\frac{P_B}{P}=\frac{P_B^{\circ }{x}_B}{P}$
Where, $P_B=\text{Partial pressure of vapour B}$
$x_B=\text{Mole fraction of B}$
The ratio of $y_A$ to $y_B$ is $\frac{y_A}{y_B}=\frac{P_A}{P_B}=\frac{P_A^{\circ }{x}_A}{P_B^{\circ }{x}_B}\Rightarrow \frac{4}{3}=\frac{1}{3}\times \frac{x_A}{(1-x_A)}$
$\Rightarrow 4=\frac{x_A}{(1-x_A)}$
$\Rightarrow x_A=\frac{4}{5}\therefore x_B=1-x_A=1-\frac{4}{5}=\frac{1}{5}$