Solutions: Numerical Problems Set (5)

Question

The vapour pressure of pure water at $25°\text{C}$ is $23.8\,\text{mm Hg}$. When $5\,\text{g}$ of a non-volatile solute is dissolved in $90\,\text{g}$ of water, the vapour pressure drops to $23.5\,\text{mm Hg}$. Find the molar mass of the solute.

Solution — Step by Step

The molar mass of the solute is approximately $79\,\text{g/mol}$.

Why This Works

Raoult’s law states that the partial pressure of solvent over a solution equals the mole fraction of solvent times the pure solvent’s vapour pressure. For non-volatile solutes, the relative lowering of vapour pressure equals the mole fraction of solute. This is a colligative property — depends only on count of solute particles, not their identity.

For the dilute approximation, mole fraction simplifies to $n_{\text{solute}}/n_{\text{solvent}}$, valid when solute is much less than solvent.

Alternative Method

Use the working formula $M_{\text{solute}} = \dfrac{w_{\text{solute}} \cdot M_{\text{solvent}}}{w_{\text{solvent}}}\cdot \dfrac{P^0}{P^0 - P}$:

$M = \frac{5 \times 18}{90} \times \frac{23.8}{0.3} = 1 \times 79.33 = 79.3\,\text{g/mol}$

Cleaner.

Common Mistake

Using molality instead of mole fraction for vapour pressure problems. Vapour pressure obeys mole fraction (Raoult); freezing point and boiling point obey molality. Different concentration units, different formulas.