Question

The vapour pressure of two pure liquids $A$ and $B$, that form an ideal solution 100 and 900 mm Hg respectively at temperature T. This liquid solution of A and B is composed of 1 mole of A and 1 mole of B.

What will be the pressure, when 1 mole of the mixture has been vaporized?

• A. 400 mm of Hg
• B. 723 mm of Hg
• C. 150 mm of Hg
• D. 300 mm of Hg

Answer 1

Answer: (D)

300 mm of Hg

Let $n_B$is the number of moles of $B$ present in 1 mole of mixture that has been vaporised

So $Y_B=\frac{n_B}{1}$

We know mole fraction is $X_B=\frac{P-P_T^o}{P_B^o-P_T^o}$

$[\because P=P_T^o+(P_B^o-P_T^o\left)R_B\right]$

Substitute the value of $X_B$ & $Y_B$ in

$1-n_B=\frac{P-P_T^o}{P_B^o-P_T^o}$ & $n_B=\frac{(1-n_B)P_B^o}{P}$

(or) $n_B=\frac{P_B^o}{P+P_B^o}\Rightarrow 1-\frac{P_B^o}{P+P_B^o}=\frac{P-P_T^o}{P_B^o-P_T^o}$

$P=\sqrt{P_B^o.P_T^o}=\sqrt{100\times 900}$

$\therefore P=300$ mm Hg

Hence, the correct option is $\text{D}$