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

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 ($x_{solvent} + x_{solute} = 1$). $x_{solvent} = 1 - x_{solute,2}$ $x_{solvent} = 1 - 0.4 = 0.6$