The vapour pressure of acetone at 20^oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20^o C, its vapour pressure was 183 torr. The molar mass (g mol^-1) of the substance is

Given, p^o = 185 torr at 20^oC and p_s - 183 torr at 20^oC Mass of non-volatile substance, m - 1.2 g 
Mass of acetone taken = 100 g and M = ?
As, we have $\frac{{\mathrm{p}}_{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}_{\mathrm{s}}}=\frac{\mathrm{n}}{\mathrm{N}}$
Putting the values, we get, $\frac{185-183}{183}=\frac{{\displaystyle \frac{12}{\mathrm{M}}}}{100} \Rightarrow \frac{2}{183}=\frac{1.2\times 58}{100\times \mathrm{M}}$
                                $\therefore \mathrm{M}=\frac{183\times 1.2\times 58}{2\times 100}=63.684 \mathrm{g} \approx 64 \mathrm{g} {\mathrm{mol}}^{-1}$