Question

Two liquids $A$ and $B$ have $P_A^0$ and $P_B^0$ in the ratio of 1 : 3 and the ratio of number of moles of $A$ and $B$ in liquid phase are 1 : 3 then mole fraction of $A$ in vapour phase in equilibrium with the solution is equal to:

• A. $0.1$
• B. $0.2$
• C. $0.5$
• D. $1.0$

Answer 1

The correct option is A $0.1$

Raoult's law : $P_1\propto x_1\Rightarrow P_1=P_1^0{x}_1$

$P_2\propto x_2\Rightarrow P_2=P_2^0{x}_2$

According to question, $\frac{P_1^0}{P_2^0}=\frac{1}{3}$ and $\frac{n_1}{n_2}=\frac{1}{3}$

$x_1=\frac{n_1}{n_1+n_2}=\frac{1}{4}$

We also know that

$x_1+x_2=1$

$x_2=1-x_1$

$P_{total}=P_1^0{x}_1+P_2^0{x}_2$

$=P_1^0{x}_1+P_2^0(1-x_1)$

$=P_1^0{x}_1+3P_1^0(1-x_1)$

$P_{total}=P_1^0(3-2x_1)⟶\left(1\right)$

to find, : $y_1=\frac{P_1}{P_{total}}=\frac{P_1^0{x}_1}{P_1^0(3-2x_1)}=\frac{x_1}{3-{2x}_1}$

$y_1=\frac{1/4}{3-2\left(1/4\right)}=\frac{1/4}{10/4}=1/10=0.1$

$\therefore$ mole fraction of $A$ is $0.1$