MHTCET 17 Apr 2026 Shift 2 Paper

Concept:For a dilute solution of a nonvolatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute, approximated as $\frac{n_2}{n_1}$.Explanation:Given: mass of solute $w_2 = 6.4\ \mathrm{g}$, mass of water $w_1 = 100\ \mathrm{g}$, molar mass of water $M_1 = 18\ \mathrm{g\ mol^{-1}}$, and relative lowering $\frac{p^0 - p}{p^0} = 0.016$.Use the formula:$$\frac{p^0 - p}{p^0} = \frac{w_2/M_2}{w_1/M_1}$$Substitute the values:$$0.016 = \frac{6.4/M_2}{100/18}$$Rearrange to solve for $M_2$:$$M_2 = \frac{6.4 \times 18}{0.016 \times 100}$$$$M_2 = \frac{115.2}{1.6} = 72\ \mathrm{g\ mol^{-1}}$$Answer:The molar mass of the solute is $72\ \mathrm{g\ mol^{-1}}$.Therefore, the correct option is C: $72\ \mathrm{g\ mol^{-1}}$.