Question

Liquids $A$ and $B$ form ideal solution over the entire range of composition. At temperature, $T$, an equimolar binary solution of liquids $A$ and $B$ has vapour pressure $45$ Torr. At the same temperature, a new solution of $A$ and $B$ having mole fractions $X_A$ and $X_B$, respectively, has a vapour pressure of $22.5$ Torr. The value of $X_A/X_B$ in the new solution is __________.

(Given that the vapour pressure of pure liquid $A$ is $20$ Torr at temperature $T$).

Answer 1

$\begin{array}{c}\text{At a certain temperature,}\\ \text{vapour pressure of equimolar binary solution}=45\text{Torr.}\\ \Rightarrow \frac{1}{2}{P}_A^{\circ }+\frac{1}{2}{P}_B^{\circ }=45\left(\text{Raoult's}Law\right)\end{array}$

$\begin{array}{c}\text{(where}{P}_A^{\circ }\text{and}{P}_B^{\circ }\text{are vapour pressure of}\\ \text{pure liquids}A\text{and}B\text{at temperature}T\text{)}\end{array}$

$\text{It is given},P_A^{\circ }=20Torr.$

$\therefore P_B^{\circ }=70\text{Torr}\text{(From eqn-(i))}$

$\begin{array}{c}\text{Now, for another solution described,}\\ \text{we may write by Raoult's Law:}\\ P_A^{\circ }{X}_A+P_B^{\circ }{X}_B=P_T=22.5\text{Torr.}\end{array}$

$\text{(where}{X}_A\text{and}{X}_B\text{are mole fraction}A\text{and}B\text{in solution)}$

$\Rightarrow 20X_A+70(1-X_A)=22.5(\because X_A+X_B=1)$

$\therefore X_A=0.95\text{and}{X}_B=0.05$

$\text{Hence,}\frac{X_A}{X_B}=\frac{0.95}{0.05}=19$