Question

Two liquids having vapor pressure $P_1^o$ and $P_2^o$ in pure state in the ratio of $2:1$ are mixed in the molar ratio of $1:2$. The ratio of their moles in the vapor state would be :

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $3:2$

Answer 1

The correct option is A $1:1$

$\frac{P_1^o}{P_2^o}=\frac{2}{1}$

Let, $P_1^o=2\&P_2^o=1$

Similarly, $n_1^o=1\&n_2^o=2$

By Raoult's law,

Vapor pressure of $1=x_1\times P_1^o=\frac{1}{3}\times 2=\frac{2}{3}$

Vapor pressure of $2=x_2\times P_2^o=\frac{2}{3}\times 1=\frac{2}{3}$

Total pressure$=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

Now,$x_1=\frac{P_1}{P_T}\&x_2=\frac{P_2}{P_T}\therefore x_{1=}\frac{2/3}{4/3}\&x_2=\frac{2/3}{4/3}\therefore x_1=1\&x_2=1$

where, $x_1$ & $x_2$ are mole fractions of $1$ & $2$ respectively.

Therefore, ratio of moles= $1:1$