Freezing Point Depression

Question

An aqueous solution of an unknown non-electrolyte solute has a freezing point of −0.744°C. Given K_f for water is 1.86 K·kg/mol and 12.0 g of the solute is dissolved in 100 g of water, what is the molar mass of the solute?

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Freezing point depression is a colligative property given by: ΔT_f = K_f × m, where m is the molality. Here, ΔT_f = 0 − (−0.744) = 0.744°C. Therefore, m = ΔT_f / K_f = 0.744 / 1.86 = 0.4 mol/kg. Since 100 g = 0.100 kg of solvent is used, moles of solute = 0.4 × 0.100 = 0.04 mol. Molar mass = mass of solute / moles = 12.0 / 0.04 = 300 g/mol. Option 100 g/mol would imply 0.12 mol of solute, giving ΔT_f = 1.86 × 1.2 = 2.23°C, far too large. Option 120 g/mol gives 0.1 mol in 0.1 kg, so m = 1.0, ΔT_f = 1.86°C, inconsistent with the data. Option 60 g/mol would give twice as many moles and too large a depression. This is a standard inverse calculation of molar mass from colligative properties — a key NCERT concept and common JEE Main question type. Plausibility check: a molar mass of 300 g/mol is in the range of small organic molecules, consistent with a non-electrolyte solute.

Freezing Point Depression

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