5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

Mole fraction of water in solution is

$x=\frac{(100/18)\mathrm{mol}}{(5 \mathrm{g}/M)+(100/18)\mathrm{mol}}$

Hence, $p=x_1{p}_1^{*}$

$2980 \mathrm{Pa}=\frac{(100/18)\mathrm{mol}}{(5 \mathrm{g}/M)+(100/18)\mathrm{mol}}(3000 \mathrm{Pa})$ This gives $M=134.1 \mathrm{g}{ \mathrm{mol}}^{-1}$