Question

The vapour pressure of a pure liquid A is $80$ mmHg at $300K$. It forms an ideal solution with liquid B. When the mole fraction of B is $0.4$, the total pressure was to be $88$mmHg. The vapour pressure of liquid B would be?

Answer 1

We know that,

$H_A{P}_A+H_B{P}_B=P$

$\Rightarrow 0.6\times 88+0.4P_B=88$

$\Rightarrow P_B=\frac{(88-0.6\times 80)}{0.4}$

$=100mmHg$ .