Question

Two liquids 'A' and 'B' have vapour pressure (in pure state) in the ratio $p_A^{\circ }:p_B^{\circ }=1:3$ at a certain temperature. Assume 'A' and 'B' form an ideal solution and the ratio of mole fractions of 'A' to 'B' in the vapour phase is 4 : 3. The the mole fraction of 'B' in the solution at the same temperature is:

• A. $\frac{2}{4}$
• B. $\frac{1}{5}$
• C. $\frac{2}{3}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{1}{5}$

Given:
$\frac{p_A^{\circ }}{p_B^{\circ }}=\frac{1}{3}$
where,
$p_A^{\circ }\Rightarrow \text{Pure vapour pressure of A}{p}_B^{\circ }\Rightarrow \text{Pure vapour pressure of B}$

$\frac{Y_A}{Y_B}=\frac{4}{3}$
where,
$Y_A\Rightarrow \text{Mole fraction of A in vapour phase}{Y}_B\Rightarrow \text{Mole fraction of B in vapour phase}$

From Dalton's law,
$p_A=P_T\times Y_A$
where,
$P_T\Rightarrow \text{Total pressure of the liquid mixture}{p}_A\Rightarrow \text{Partial vapour pressure of A}{Y}_A\Rightarrow \text{Mole fraction of A in vapour phase}$

From Raoult's law,
$p_A=p_A^{\circ }\times X_A$
where,
$p_A\Rightarrow \text{Vapour pressure of A}{X}_A\Rightarrow \text{Mole fraction of A in liquid}{p}_A^{\circ }\Rightarrow \text{Pure vapour pressure of A}$
Thus,
$Y_A=\frac{p_A}{P_T}\Rightarrow \frac{p_A^{\circ }{X}_A}{P_T}....eqn\left(i\right)$
Similarly,
$Y_B=\frac{p_B^{\circ }{X}_B}{P_T}.....eqn\left(ii\right)$
Dividing equation (i) and (ii), we get,
$\frac{Y_A}{Y_B}=\frac{p_A^{\circ }}{p_B^{\circ }}\times \frac{X_A}{X_B}\Rightarrow \frac{4}{3}=\frac{1}{3}\times \frac{X_A}{(1-X_A)}4=\frac{X_A}{(1-X_A)}{X}_A=\frac{4}{5}{X}_B=1-\frac{4}{5}{X}_B=\frac{1}{5}$