The vapour pressure of a solvent is decreased by $10\text{mm}$ of Hg, when a non volatile solute was added to the solvent. The mole fraction of the solute in the solution is $0.2$, what should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20$ mm of Hg?

The correct option is B $0.6$
Let, ${\chi }_1$ is the mole fraction of the solute
So, ${\chi }_1=0.2$
From Raolt's law,
$\frac{P^0-P_s}{P^0}=x_1$
Where, $P^o=\text{Vapour pressure of pure solvent}$
$P_{\left(s\right)}=\text{Vapour pressure of solute}$
$\Rightarrow \frac{10}{P^0}=0.2\Rightarrow P^0=50\text{mm of Hg}$
Again, when $P^0-P_s$ = $20$ mm of Hg, then
$\frac{P^0-P_s}{P^0}$= mole fraction of solute = $\frac{20}{50}=0.4$
Therefore mole fraction of solvent $=1-0.4=0.6$