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: Governing relation. For a dilute solution the fall in freezing point equals the molal depression constant times the molality: \(\Delta T_f = K_f\, m\). The molality itself equals moles of solute per kilogram of solvent.

Step 2: Moles of solute in symbols. Let \(M\) be the required molar mass. Moles of solute \(= \dfrac{0.32}{M}\). Mass of solvent \(= 25\,\text{g} = 0.025\,\text{kg}\). Hence molality \(m = \dfrac{0.32/M}{0.025} = \dfrac{12.8}{M}\).
Step 3: Insert the observed depression. The solution freezes at \(-0.201\,^\circ\text{C}\), so \(\Delta T_f = 0.201\,^\circ\text{C}\). Then \(0.201 = 1.86 \times \dfrac{12.8}{M}\).
Step 4: Solve for \(M\). \(M = \dfrac{1.86 \times 12.8}{0.201} = \dfrac{23.808}{0.201} = 118.45\).
\[\boxed{M \approx 118.4\,\text{g mol}^{-1}}\]
Both routes give the same value, confirming the molecular weight of the compound is about \(118\,\text{g mol}^{-1}\).