A solution is made by dissolving 30 g of a non-volatile solute in 90g of water. It has a vapour pressure of 2.8 kPa at 298 K. At 298 K, vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of the solute.

The relative lowering of vapour pressure is given by the following expression. 
$\frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -{\mathrm{p}}_{\mathrm{solution}}}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{n_2}{n_1+n_2}$

Where ${\mathrm{p}}_{\mathrm{solvent}}^0$ is the vapour pressure of pure solvent, P_solution  is the vapour pressure of solution containing dissolved solute, n₁ is the number of moles of solvent and n₂ is the number of moles of solute.

For dilute solution n₂<<n₁, therefore the above expression reduces to 

$$\begin{gathered} \frac{{\mathrm{p}}_{\mathrm{solvent}}^0- {\mathrm{p}}_{\mathrm{solution}}}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{n_2}{n_1} \\ = \frac{w_2 x{M}_1}{M_2 x w_1} .....\left(A\right) \end{gathered}$$

Where w₁ and w₂ are the masses and M₁ and M₂ are the molar masses of solvent and solute respectively.

We are given that

W₁ =30 g

W₂ =30g

p_solution =2.8 kpa

p⁰_solvent =? And M₂=?

$$\begin{gathered} \mathrm{subtituting} \mathrm{these} \mathrm{values} \mathrm{in} \mathrm{relation} \\ \mathrm{we} \mathrm{get}, \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{30 \mathrm{x}18}{{\mathrm{M}}_2 \mathrm{x}90} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{6}{{\mathrm{M}}_2 } \left(1\right) \\ \mathrm{similarly} \mathrm{for} \mathrm{second} \mathrm{case} \mathrm{we} \mathrm{have} \mathrm{the} \mathrm{following} \mathrm{value}. \\ {\mathrm{w}}_2 =90 \mathrm{g} \\ {\mathrm{w}}_1 =90+18 =108\mathrm{g} \\ {\mathrm{p}}_{\mathrm{solution}} =2.9 \mathrm{kpa} \\ \mathrm{therefore} \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.9}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{30 \mathrm{x}18}{{\mathrm{M}}_2 \mathrm{x}108} \left(2\right) \\ \mathrm{dividing} 1 \mathrm{and} 2 \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0-2.9} =\frac{6}{5} \\ \mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{water} \mathrm{at} 298 \mathrm{k} \mathrm{is} 3.4\mathrm{kpa} \\ \mathrm{substituting} \mathrm{the} \mathrm{vaue} \mathrm{of} {\mathrm{p}}_{\mathrm{solvent}}^0 \mathrm{in} 1 \mathrm{we} \mathrm{get} \\ \frac{3.4 -2.8}{3.4} =\frac{6}{{\mathrm{M}}_2} \\ = 34 \mathrm{gram} \\ \mathrm{Therefore} \mathrm{mass} \mathrm{of} \mathrm{solute} \mathrm{is} 34 \mathrm{g} . \end{gathered}$$