Question

The vapour pressure of two pure liquids 'A' and 'B', that form an ideal solution are $100$ and $900torr$ respectively at temperature, T. This liquid solution of 'A' and 'B' is composed of $1mol$ of A and $1mol$ of B. What will be the total vapour pressure of mixture, when $1mol$ of mixture has been vaporised?

• A. $800torr$
• B. $500torr$
• C. $300torr$
• D. None of these

Answer 1

The correct option is C $300torr$

Let, $n_B$ mole of B present in 1 mole of mixture that has been vaporized. Thus, $\text{Mole fraction of B in vapour,}{Y}_B=\frac{n_B}{1}$
Mole fraction of B in the remaining liquid phase will be
$X_B=\frac{1-n_B}{1}$
By Raoult's law,
$X_B=\frac{P_T-p_A^{\circ }}{p_B^{\circ }-p_A^{\circ }}....eqn\left(i\right)[\because P_T=p_A^{\circ }+(p_B^{\circ }-p_A^{\circ }\left)X_B\right]$
where,
$p_A^{\circ }\Rightarrow \text{Pure vapour pressure of A}{p}_B^{\circ }\Rightarrow \text{Pure vapour pressure of B}$
$P_T\Rightarrow \text{Total pressure of the liquid mixture}$
and $Y_B=\frac{p_B}{P_T}\Rightarrow \frac{p_B^{\circ }{X}_B}{P_T}....eqn\left(ii\right)\left(\text{By Raoults law}\right)$
After substitution of values of $X_B$ and $Y_B$ in equation (1) and (2)
we get,
$1-n_B=\frac{P_T-p_A^{\circ }}{p_B^{\circ }-p_A^{\circ }}....eqn\left(iii\right)\text{and}{n}_B=\frac{(1-n_B)p_B^{\circ }}{P_T}....eqn\left(iv\right)$
Further solving (iv)
$n_B\times P_T=p_B^{\circ }-n_B{p}_B^{\circ }{n}_B.P_T+n_B.p_B^{\circ }=p_B^{\circ }$
$n_B=\frac{p_B^{\circ }}{P_T+p_B^{\circ }}....eqn\left(v\right)$
Substituting equation $\left(v\right)$ in $\left(iii\right)$
we get,
$1-\frac{p_B^{\circ }}{P_T+p_B^{\circ }}=\frac{P_T-p_A^{\circ }}{p_B^{\circ }-p_A^{\circ }}\frac{P_T}{P_T+P_B^{\circ }}=\frac{P_T-p_A^{\circ }}{p_B^{\circ }-p_A^{\circ }}{P}_T.p_B^{\circ }-P_T.p_A^{\circ }=P_T^2-P_T.p_A^{\circ }+P_T.p_B^{\circ }-p_B^{\circ }.p_A^{\circ }\Rightarrow P_T=\sqrt{p_B^{\circ }.p_A^{\circ }}=\sqrt{900\times 100}\Rightarrow P_T=300torr$