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

Vapour pressure of pure acetone, P0A = 185 torr
Vapour pressure of solution, P = 183 torr
Molar mass of solvent, M_A = 58 g mol^-1
W_B = 1.2 g, W_A = 100 g
$\begin{array}{l}\frac{{\mathrm{P}}^0\mathrm{A}-\mathrm{P}}{{\mathrm{P}}^0\mathrm{A}}={\mathrm{x}}_{\mathrm{B}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}\\ \frac{{\mathrm{P}}^0\mathrm{A}-\mathrm{P}}{{\mathrm{P}}^0\mathrm{A}}=\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\times \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{A}}}\\ \frac{185-183}{185}=\frac{1.2}{{\mathrm{M}}_{\mathrm{B}}}\times \frac{58}{100}\\ \frac{2}{185}=\frac{1.5\times 58}{{\mathrm{M}}_{\mathrm{B}}\times 100}\\ \therefore {\mathrm{M}}_{\mathrm{B}}=\frac{1.2\times 58\times 185}{2\times 100}=64.38\end{array}$