When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution:
- First increases, then decreases
- Decreases
- Increases
- Remains the same
Show answer and solution
Correct answer: B. Decreases
The non-volatile solute occupies some of the surface that would otherwise emit solvent vapour, reducing the rate of vaporization.
**Hence $P_{solution} < P^0_{solvent}$ — Raoult's law gives $P = x_1 P_1^0$ with $x_1 < 1$.**