Question

18 g of a non-volatile solute is dissolved in 200 g of H$_2$O, and it freezes at 272.07 K. Calculate the molecular mass of the solute (K$_f$ for water = 1.86 K kg mol$^{-1}$).

Hint:

Freezing point depression is a colligative property that helps determine the molecular mass of solutes by observing changes in freezing point.

Answer 1

We are given:
- Mass of solute, \( w_B = 18 \, \text{g} \)
- Mass of solvent, \( w_A = 200 \, \text{g} \)
- Freezing point depression, \( \Delta T_f = T^\circ_f - T_f = 273.15 \, \text{K} - 272.07 \, \text{K} = 1.08 \, \text{K} \)
- Freezing point depression constant for water, \( K_f = 1.86 \, \text{K kg mol}^{-1} \)
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot m \] Where \( m \) is the molality, defined as \( m = \frac{w_B \times 1000}{M_B \times w_A} \), and \( M_B \) is the molar mass of the solute. Replacing values:
\[ 1.08 = 1.86 \times \frac{18 \times 1000}{M_B \times 200} \] Solving for \( M_B \):
\[ M_B = \frac{1.86 \times 18 \times 1000}{200 \times 1.08} = 155 \, \text{g mol}^{-1} \]