Molar Mass of Solute from Vapour Pressure Lowering

We studied that the relative lowering of vapour pressure is equal to the mole fraction $x_2$ of the solute in the solution: from Eq. (2.6) it follows that $\dfrac{\Delta P}{P_1^0} = x_2$ (Eq. 2.8). Recall that the mole fraction of a component of a solution is equal to its moles divided by the total moles in the solution. Thus,

where $n_1$ and $n_2$ are the moles of solvent and solute respectively, in the solution.

We are concerned only with dilute solutions, hence $n_1 \gg n_2$ and $n_1 + n_2 \approx n_1$. The mole fraction $x_2$ is then given by

Suppose that we prepare a solution by dissolving $\mathrm{W_2}$ g of a solute in $\mathrm{W_1}$ g of solvent. The moles of solute and solvent in the solution are then

where $M_1$ and $M_2$ are the molar masses of solvent and solute, respectively. Substitution of Eq. (2.10) into Eq. (2.9) yields …