Prove that the relative lowering of vapour pressure is equal to the mole fraction of non-volatile solute in the solution.

Answer:

Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure of
the solution, mole fraction and vapour pressure of the solvent, i.e.,

$p_1= x_1{p}_1^0$         .................1

The reduction in the vapour pressure of solvent (Δp₁) is given as:

$$\begin{gathered} ∆{\mathrm{p}}_1 = {\mathrm{p}}_1^0 -{\mathrm{p}}_1 = {\mathrm{p}}_1^0 - {\mathrm{p}}_1^0{\mathrm{x}}_1 \\ = {\mathrm{p}}_1^0 (1-{\mathrm{x}}_1) ...................... \left(2\right) \\ \mathrm{Knowing} \mathrm{that} {\mathrm{x}}_2 = 1 - {\mathrm{x}}_1 \mathrm{equation} 2 \mathrm{reduces} \mathrm{to} \\ ∆p_1 = x_2{p}_1^0 ........................\left(3\right) \end{gathered}$$

In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Equation 3 can be written as :
$\frac{∆p_1}{p_1^0} = \frac{p_1^0 - p_1}{p_1^0} = x_2$

The expression on the left hand side of the equation as mentioned

earlier is called relative lowering of vapour pressure and is equal to
the mole fraction of the solute. The above equation can be written as:

$\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1+n_2} \left(\sin ce x_2\frac{n_2}{n_1+n_2}\right)$

Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 < < n1, hence neglecting n2 in the denominator we have

$\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1}$
hence proved