According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($x_{solute}$).

Formula: $\frac{\Delta p}{p^0} = x_{solute}$ Where $\Delta p$ is the lowering of vapour pressure and $p^0$ is the vapour pressure of the pure solvent.

Step 1: Calculate the vapour pressure of the pure solvent ($p^0$). Given: $\Delta p_1 = 10 \text{ mm Hg}$ $x_{solute,1} = 0.2$

Substituting into the formula : $\frac{10}{p^0} = 0.2$ $p^0 = \frac{10}{0.2} = 50 \text{ mm Hg}$

Step 2: Calculate the mole fraction of the solute for the second case. Given: $\Delta p_2 = 20 \text{ mm Hg}$ $p^0 = 50 \text{ mm Hg}$

$\frac{20}{50} = x_{solute,2}$ $x_{solute,2} = 0.4$

Step 3: Calculate the mole fraction of the solvent. The sum of mole fractions of components in a binary solution is unity ($x_{solvent} + x_{solute} = 1$) . $x_{solvent} = 1 - x_{solute,2}$ $x_{solvent} = 1 - 0.4 = 0.6$

The mole fraction of the solvent is 0.6.