Elevation of Boiling Point

A liquid boils, at any given external pressure, at exactly the temperature at which its vapour pressure becomes equal to that external pressure. Dissolving a non-volatile solute lowers the solvent's vapour pressure at every temperature (as shown by the relative-lowering relation of the previous section), so the solution's vapour pressure curve, plotted against temperature, lies everywhere below the pure solvent's curve. Consequently, the solution's vapour pressure only reaches the external pressure at a higher temperature than the pure solvent needs — the boiling point of the solution is elevated above that of the pure solvent.

Experimentally, this elevation of boiling point, $\Delta T_b = T_b(\text{solution}) - T_b^{\circ}(\text{pure solvent})$, is found to be directly proportional to the molal concentration ($m$, moles of solute per kilogram of solvent) of the solution, for reasonably dilute solutions:

The proportionality constant $K_b$ is the molal elevation constant, or ebullioscopic constant, of the solvent — it depends only on the identity of the solvent (through properties such as its normal boiling point and its enthalpy of vaporization), never on the solute, and carries units of $\text{K kg mol}^{-1}$ (equivalently $^{\circ}\text{C kg mol}^{-1}$). Water has $K_b = 0.52\ \text{K kg mol}^{-1}$; other common solvents have considerably larger constants — benzene, for instance, has $K_b = 2.53\ \text{K kg mol}^{-1}$ — reflecting differences in how sensitively each solvent's vapour pressure curve responds to a given change in the number of dissolved particles.

Because $\Delta T_b$ can be measured with a sufficiently precise thermometer, and $K_b$ is known for the solvent used, this relation gives a direct experimental route to an unknown solute's molar mass. If $w_2$ grams of solute is dissolved in $w_1$ grams of solvent, the molality is $m = (w_2/M_2) \times (1000/w_1)$, and substituting into $\Delta T_b = K_b m$ and rearranging for $M_2$ gives …