Raoult's Law and Vapour Pressure of Solutions

For a solution of two volatile components (A and B), each component obeys Raoult's law independently:

$p_A = p_A^\circ \cdot x_A \qquad p_B = p_B^\circ \cdot x_B$

By Dalton's law of partial pressures, the total vapour pressure:
$p_{\text{total}} = p_A + p_B = p_A^\circ x_A + p_B^\circ x_B$

where $p_A^\circ$ and $p_B^\circ$ are the vapour pressures of pure A and pure B, and $x_A$, $x_B$ are their mole fractions in solution.

Key result: $p_{\text{total}}$ lies between $p_A^\circ$ and $p_B^\circ$ — the total pressure is always intermediate.

When the solute is non-volatile (solid dissolved in liquid — e.g. NaCl in water), it contributes zero vapour pressure. Only the solvent evaporates:

$p_{\text{solution}} = p_1^\circ \cdot x_1$

Since $x_1 = 1 - x_2$ (mole fraction of solvent = 1 − mole fraction of solute):
$p_{\text{solution}} = p_1^\circ (1 - x_2)$

Relative lowering of vapour pressure (RLVP):
$\frac{p_1^\circ - p_{\text{solution}}}{p_1^\circ} = x_2$

RLVP equals the mole fraction of the solute — a colligative property that depends only on how many solute particles are present, not what they are.