Question

0.32 g of a new compound was dissolved in 25 g of water. The freezing point of this solution was found to be \(-0.201\,^\circ\text{C}\). Determine the molecular weight of this new compound. For water \(K_f = 1.86\,^\circ\text{C}\;\text{kg mol}^{-1}\).

Hint:

Use \(\Delta T_f = K_f\, m\) with \(\Delta T_f = 0.201\,^\circ\text{C}\), then write molality in terms of the unknown molar mass and solve.

Answer 1

Step 1: Concept. Depression in freezing point is a colligative property. \[ \Delta T_f = K_f \times m \] where \(m\) is the molality of the solution.

Step 2: Find the depression in freezing point. Pure water freezes at \(0\,^\circ\text{C}\), and the solution freezes at \(-0.201\,^\circ\text{C}\). \[ \Delta T_f = 0 - (-0.201) = 0.201\,^\circ\text{C} \]
Step 3: Write molality in terms of the unknown molar mass. With mass of solute \(w_2 = 0.32\,\text{g}\), mass of solvent \(w_1 = 25\,\text{g}\) and molar mass \(M_2\), \[ m = \frac{w_2 \times 1000}{M_2 \times w_1} \] so \[ \Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} \]
Step 4: Rearrange for \(M_2\) and substitute. \[ M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} = \frac{1.86 \times 0.32 \times 1000}{0.201 \times 25} \]
Step 5: Arithmetic. Numerator \(= 1.86 \times 0.32 \times 1000 = 595.2\). Denominator \(= 0.201 \times 25 = 5.025\). \[ M_2 = \frac{595.2}{5.025} = 118.45\,\text{g mol}^{-1} \]
\[\boxed{M_2 \approx 118.4\,\text{g mol}^{-1}}\]