Question

Two liquids having vapour pressures $P_1^{\circ }$ and $P_2^{\circ }$ 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 vapour state would be

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

Answer 1

Answer: (A)

$P_1^{\circ }:P_2^{\circ }=\frac{2}{1}$
$\frac{n_1}{n_2}=\frac{1}{2}$
$\therefore P_1=P_0^1{x}_1$
$P_2=P_2^{\circ }{x}_2$
$x_1=\frac{n_1}{n_1+n_2}=\frac{1}{1+2}=\frac{1}{3}$
$P_1^0{x}_1;P_2^0{x}_2$
$\Rightarrow 2\times \frac{1}{3};1\times \frac{2}{3}$
$\Rightarrow \frac{2}{3};\frac{2}{3}$
$x_2=\frac{n_2}{n_1+n_2}=\frac{2}{1+3}=\frac{2}{3}$
$\therefore P_1^0{x}_1=\frac{2}{3}$
$P_2^0{x}_2=\frac{2}{3}\rfloor$
$\frac{y_1}{y_2}=\frac{P_1\times P_{\text{total }}}{P_{\text{total }}\times P_2}$
$y=$ mole fraction in vapor phase
$\Rightarrow \frac{P_1^0{x}_1}{P_2^0{x}_2}\Rightarrow \frac{\frac{2}{3}}{\frac{2}{3}}$
$\Rightarrow 1:1$