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

Answer: (B)

Given, $p_A:P_B=1:2$

$\therefore p_B=2p_A$

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, $p_T=P_A{x}_A+p_B{x}_B$

$=p_A\times \frac{1}{3}+2p_A\times \frac{2}{3}$

$=\frac{1}{3}{p}_A+\frac{4}{3}{p}_A$

$=\frac{5}{3}{p}_A$

Mole fraction in vapour phase

$y_A=\frac{p_A{x}_A}{pr}$

$=\frac{p_A\times \frac{1}{3}}{\frac{5}{3}{p}_A}=\frac{1}{5}=0.2$