Relative Lowering of Vapour Pressure

The vapour pressure of a solution containing a nonvolatile, non-electrolyte solute is always lower than the vapour pressure of the pure solvent alone (this was already derived from Raoult's law in section 9.7.2). Here is the thermodynamic reasoning for why: consider a closed system in which the pure solvent liquid is in equilibrium with its own vapour -- at that equilibrium, the molar Gibbs free energies of the liquid and vapour phases are equal, $\Delta G=0$. Now dissolve a solute into the liquid: the dissolution process increases entropy, which lowers the liquid's free energy ($G$). To restore the liquid-vapour equilibrium, the vapour phase's free energy must fall by a matching amount -- and at a fixed temperature, the only way to lower a gas's free energy is to lower its pressure. So the solution's vapour pressure must decrease relative to the pure solvent's, in order for equilibrium to be maintained (Figure 9.10).

From section 9.7.2 (equation 9.16), Raoult's law shows this relative lowering of vapour pressure is numerically equal to the solute's mole fraction, and is therefore independent of the solute's chemical identity -- exactly the signature of a colligative property.

Determining molar mass from relative lowering of vapour pressure. This measurement can be turned around to find an unknown solute's molar mass. Let $w_A$ g of solvent (molar mass $M_A$) hold $w_B$ g of dissolved solute (molar mass $M_B$). The solute's exact mole fraction is

For a dilute solution, $n_A\gg n_B$, so $n_A+n_B\approx n_A$ and this simplifies to $x_B\approx\dfrac{n_B}{n_A}$. Substituting $n_A=w_A/M_A$ and $n_B=w_B/M_B$ (eq. 9.21):

Once the relative lowering of vapour pressure is measured experimentally, and $w_A$, $w_B$ and $M_A$ (the solvent's known molar mass) are known, this equation can be solved for the unknown $M_B$. …