Vapour Pressure Lowering

When a liquid in a closed container is in equilibrium with its vapours, the pressure exerted by the vapour on the liquid is its vapour pressure.

i. Experiments have shown that when a nonvolatile, nonionizable solid is dissolved in a liquid solvent, the vapour pressure of the solution is lower than that of the pure solvent. In other words, the vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it. When the solute is nonvolatile it does not contribute to the vapour above the solution; therefore the vapour pressure of the solution is equal to the vapour pressure of the solvent above the solution.

ii. If $P_1^0$ is the vapour pressure of the pure solvent and $P_1$ is the vapour pressure of the solvent above the solution, then $P_1 < P_1^0$. The vapour pressure lowering is

$\Delta P = P_1^0 - P_1 \qquad \text{...(2.5)}$ …