According to Raoult's Law for solutions containing non-volatile solutes, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($x_{solute}$). The formula is given by: $\frac{\Delta P}{P^0} = x_{solute}$ 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 ($P^0$). Using the first set of data: $\Delta P_1 = 10 \text{ mm Hg}$ $x_{solute,1} = 0.2$ 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 new condition. Using the second set of data: $\Delta P_2 = 20 \text{ mm Hg}$ $P^0 = 50 \text{ mm Hg}$ (constant for the solvent at a given temperature) $\frac{20}{50} = x_{solute,2}$ $x_{solute,2} = 0.4$

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