Question

An aqueous solution containing 0.2 g of a non-volatile solute 'A' in 21.5 g of water freezes at 272.814 K. If the freezing point of water is 273.16 K, the molar mass (in g mol$^{-1}$) of solute A is [K$_f$(H$_2$O) = 1.86 K kg mol$^{-1}$]

• A. 80
• B. 75
• C. 100
• D. 50

Hint:

$\Delta T_f = K_f m$. Molality (m) = moles of solute / kg of solvent.

Answer 1

The Correct Option is D

Freezing point depression ($\Delta T_f$) is given by $\Delta T_f = K_f m$, where $K_f$ is the molal freezing point depression constant and $m$ is the molality. $$ \Delta T_f = 273.16 \, \text{K} - 272.814 \, \text{K} = 0.346 \, \text{K} $$ $$ m = \frac{\Delta T_f}{K_f} = \frac{0.346 \, \text{K}}{1.86 \, \text{K kg mol}^{-1}} \approx 0.186 \, \text{mol kg}^{-1} $$ Molality is defined as moles of solute per kilogram of solvent. Let $M$ be the molar mass of solute A. $$ 0.186 \, \text{mol kg}^{-1} = \frac{0.2 \, \text{g} / M}{21.5 \, \text{g} \times 10^{-3} \, \text{kg/g}} $$ $$ M = \frac{0.2}{0.186 \times 0.0215} \approx 50 \, \text{g/mol} $$