Question

An ideal solution was found to have a vapour pressure of 180 torr when the mole fraction of a non-volatile solute was 0.1. What would be the vapour pressure of the pure solvent (in torr) at the same temperature?

Answer 1

200

Answer 2

Raoult's Law states that for a solution containing a non-volatile solute, the vapour pressure of the solution ($P_{solution}$) is directly proportional to the mole fraction of the solvent ($X_{solvent}$). The proportionality constant is the vapour pressure of the pure solvent ($P^0_{solvent}$). $P_{solution} = X_{solvent} \times P^0_{solvent}$ The mole fraction of the solvent can be calculated from the mole fraction of the solute ($X_{solute}$) using the relationship: $X_{solvent} = 1 - X_{solute}$ Given: Vapour pressure of the solution, $P_{solution} = 180$ torr. Mole fraction of the non-volatile solute, $X_{solute} = 0.1$.

First, calculate the mole fraction of the solvent: $X_{solvent} = 1 - 0.1 = 0.9$ Now, apply Raoult's Law to find the vapour pressure of the pure solvent ($P^0_{solvent}$): $180 \text{ torr} = 0.9 \times P^0_{solvent}$ Solving for $P^0_{solvent}$: $P^0_{solvent} = \frac{180 \text{ torr}}{0.9}$ $P^0_{solvent} = \frac{1800}{9} \text{ torr}$ $P^0_{solvent} = 200 \text{ torr}$