The vapour pressure of acetone at 20 °C is 185 Torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 °C, its vapour pressure was 193 Torr. The molar mass of the substance is

According to Raoult's law,

$p_1=x_1{p}_1^{\ast }=\left(1-x_2\right)p_1^{\ast }\text{ i.e. }{x}_2=1-\frac{p_1}{p_1^{\ast }}=1-\frac{183\text{ Torr }}{185\text{ Torr }}=\frac{2}{185}$Also,    $x_2=\frac{n_2}{n_1+n_2}=\frac{(1.2\mathrm{g}/M)}{\left(100\mathrm{g}/58\mathrm{g}{\mathrm{mol}}^{-1}\right)+(1.2\mathrm{g}/M)}$

Hence,   $\frac{(1.2\mathrm{g}/M)}{\left(100\mathrm{g}/58\mathrm{g}{\mathrm{mol}}^{-1}\right)+(1.2\mathrm{g}/M)}=\frac{2}{185}$

This gives

$\frac{1.2\mathrm{g}}{M}\left(1-\frac{2}{185}\right)=\left(\frac{2}{185}\right)\left(\frac{100\mathrm{g}}{58\mathrm{g}{\mathrm{mol}}^{-1}}\right)$  or  $M=\left(1.2\mathrm{g}\right)\left(\frac{183}{185}\right)\left(\frac{185}{2}\right)\left(\frac{58\mathrm{g}{\mathrm{mol}}^{-1}}{100\mathrm{g}}\right)=63.68\mathrm{g}{\mathrm{mol}}^{-1}$

Approixmate value Writing $x_2\simeq n_2/n_1\text{, we get }\frac{2}{185}=\frac{1.2\mathrm{g}/\mathrm{M}}{\left(100\mathrm{g}/58\mathrm{g}{\mathrm{mol}}^{-1}\right)}\text{. }$

This gives $M=\left(1.2\times \frac{58}{100}\times \frac{185}{2}\right)\mathrm{g}{\mathrm{mol}}^{-1}=64.38\mathrm{g}{\mathrm{mol}}^{-1}$