Question

Two liquids A and B have vapour pressures in the ratio $P_A^0:P_B^0=2:3$ at a certain temperature. Assuming that A and B form an ideal solution and ratio of mole fractions of A and B in the vapour phase is 1 : 3 , then the mole fraction of A in the solution at the same temperature is :

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{3}$

Given, Vapour pressure rate of two liquids $A$ & $B$

$P_A^{\circ }:P_B^{\circ }=2:3$

$\therefore$ Total vapour pressure=$2+3=5$

Also given $A$ & $B$ from ideal solution

& Mole fraction of $A$ & $B$ form ratio $1:3$ in vapour phase

$\Rightarrow$ Mole fraction of $A$=$\frac{1}{1+13}$ & Mole fraction of $B=\frac{3}{1+3}$

$=\frac{1}{4}$ $=\frac{3}{4}$

According to Dalton's law of partial pressure & Raolt's law

$P_A=X_A\times P_{Total}$ $P_A=$ Partial pressure of $A$

$=\frac{1}{4}\times 5$ $X_A=$ Mole factor of $A$

$=\frac{5}{4}$

Same,

$P_B=X_B\times P_{Total}$ $P_B=$ partial pressure of $B$

$=\frac{3}{4}\times 5$ $X_B$=Mole fraction of $B$

Thus, mole fraction of $A$ in solution phase $X_A$ will be as $X_A\propto \frac{P_A}{P_B}$$\Rightarrow X_A=\frac{5}{4}:\frac{15}{4}$

$=\frac{1}{3}$

Hence, the correct option is $A$