Relative Lowering of Vapour Pressure

The vapour pressure of a solvent in a solution is always less than that of the pure solvent. Raoult found that this lowering depends only on the concentration of the solute particles and not on their identity — the first sign that we are dealing with a colligative property.

From Raoult's law to the lowering

For a solution containing a non-volatile solute, Raoult's law gives the vapour pressure of the solvent as

where $p_1$ is the vapour pressure of the solvent over the solution, $p_1^{0}$ is the vapour pressure of the pure solvent, and $x_1$ is the mole fraction of the solvent.

The reduction in the solvent's vapour pressure, written $\Delta p_1$, is the difference between the pure-solvent value and the solution value:

Since a solution contains only solvent and solute, $x_1 + x_2 = 1$, so $1 - x_1 = x_2$. Substituting:

Here $x_2$ is the mole fraction of the solute.

Relative lowering of vapour pressure

Dividing the lowering by the pure-solvent vapour pressure gives a quantity that equals the mole fraction of the solute directly:

Extending to moles, then to masses

Writing $x_2$ in terms of the numbers of moles of solvent ($n_1$) and solute ($n_2$):

For dilute solutions the solute is in very small amount, so $n_2 \ll n_1$. Neglecting $n_2$ in the denominator:

Now express the moles through masses and molar masses, using $n_1 = w_1/M_1$ and $n_2 = w_2/M_2$:

$\frac{p_1^{0} - p_1}{p_1^{0}} = \frac{w_2 \times M_1}{M_2 \times w_1}$ …