Question

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1 : 2$. A binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1 : 2$. The mole fraction of $A$ in the vapour phase of the solution will be

• A. $0.33$
• B. $0.2$
• C. $0.25$
• D. $0.52$

Answer 1

The Correct Option is B

Given, $p _{ A }: P _{ B }=1: 2$
$\therefore p _{ B }=2 p _{ A }$

Similarly $\frac{n_{A}}{n_{B}}=\frac{1}{2}$
$\therefore x_{A}=\frac{1}{1+2}=\frac{1}{3}$ and $x_{B}=\frac{2}{1+2}=\frac{2}{3}$

Total pressure, $p _{ T }= P _{ A } x _{ A }+ p _{ B } x _{ B }$
$=p_{A} \times \frac{1}{3}+2 p_{A} \times \frac{2}{3}$
$=\frac{1}{3} p_{A}+\frac{4}{3} p_{A}$
$=\frac{5}{3} p_{A}$

Mole fraction in vapour phase

$y_{A}=\frac{p_{A} x_{A}}{p r}$
$=\frac{p_{A} \times \frac{1}{3}}{\frac{5}{3} p_{A}}=\frac{1}{5}=0.2$