The vapour pressure of a solvent decreased by $10mm$ of $Hg$ when a non-volatile solute was added to the solvent. The mole fraction of solute in the solution is $0.2$. What would be the mole fraction of solvent if decrease in vapour pressure is $20mm$?

The correct option is C $0.6$
$p^0=$ vapour pressure of pure solvent

$p^s=$ vapour pressure of of solvent in soultion

$x_2=$mole fraction of solute in solution

$x_1$=1-$x_2=$ mole fraction of solvent in solution

As we know that, ${(p}^0-p^s)/p^0=x_2$

Case 1: Decrease in vapour pressure= 10mm of hg

$10/$ $p^0=0.2$

$p^0=50$ mm of hg

Case 2: Decrease in vapor pressure=20mm of hg

we find $p^0=50$ mm of hg from case 1

$20/50$ = { x }_{ 2 }$

$x_2=0.4$

$x_1$=1-$x_2=0.6$

So, the correct option is $B$