0.5 g of nonvolatile solute is dissolved in 100 g of ethyl acetate in 20C.the vapour pressure of the solution and pure ethyl acetate are 72 and 72.8 torr rezpectively at 20C. Calculate the molecular weight of the solute

Dear student,

Please find below the solution to the query posted by you.

We are given,

Mass of non-volatile solute, m₁ = 0.5 g
Mass of solvent (ethyl acetate, CH_​3COOCH₂CH₃), m₂ = 100 g
Molar mass of solvent, M₂ = 88 g
T = 20 ^oC = 293 K
Vapour pressure of pure ethyl acetate, p₂^o = 72.8 torr
Vapour pressure of solution, p_s = 72 torr
Mole fraction of the solute can be determined by using the formula for relative lowering of vapour pressure$$\begin{gathered} \frac{{\mathrm{p}}_2^{\mathrm{o}} - {\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}_2^{\mathrm{o}}}=\frac{{\mathrm{m}}_1 \times {\mathrm{M}}_2}{{\mathrm{M}}_1 \times {\mathrm{m}}_2} \\ \frac{72.8 - 72}{72.8} = \frac{0.5 \times 88}{{\mathrm{M}}_1 \times 100} \\ {\mathrm{M}}_1 = \frac{0.5 \times 88 \times 72.8}{0.8 \times 100}= 40.04 \mathrm{g} {\mathrm{mol}}^{-1} \end{gathered}$$.

Hence, the molar mass of the non-volatile solute is 40.04 g mol⁻¹.

Hope this information will clear your doubt regarding relative lowering of vapour pressure.

In case of any doubt just post your query on the forum and our experts will try to help you as soon as possible.

Keep Learning!