The molecular mass of the solute ($M_2$) can be calculated using the formula for depression of freezing point: $M_2 = \frac{1000 \times w_2 \times K_f}{\Delta T_f \times w_1}$

Where: $w_2$ = mass of solute = $0.1 \text{ g}$ $w_1$ = mass of solvent = $21.7 \text{ g}$ $K_f$ = molal freezing point depression constant = $1.86 \text{ K kg mol}^{-1}$ $\Delta T_f$ = depression in freezing point = $T_f^0 - T_f$ To match the options, we assume the freezing point of pure water ($T_f^0$) is taken as $273 \text{ K}$ (an approximation often used in older problems). $\Delta T_f = 273 \text{ K} - 272.817 \text{ K} = 0.183 \text{ K}$

Substituting the values: $M_2 = \frac{1000 \times 0.1 \times 1.86}{0.183 \times 21.7}$ $M_2 = \frac{186}{3.9711}$ $M_2 \approx 46.83 \text{ g mol}^{-1}$