A 2.154 g sample of an unknown non-electrolyte is dissolved in 50 g of water (K_f of water = 1.86 °C). If the solution freezes at -1.23 °C, the molecular mass of the non-electrolyte is ________.

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Quizplus · Accepted Answer

The freezing point depression formula is ΔT_f = i * K_f * m. Since the solute is a non-electrolyte, i = 1. ΔT_f = 1.23°C, K_f = 1.86°C kg/mol. Solving for molality (m), m = ΔT_f / K_f = 1.23 / 1.86 = 0.6613 mol/kg. The mass of the solvent is 0.05 kg, so moles of solute = 0.6613 mol/kg * 0.05 kg = 0.033065 mol. The molecular mass = mass of solute / moles of solute = 2.154 g / 0.033065 mol = 65.1 g/mol, which is closest to 65.

A 2154 G Sample of an Unknown Non-Electrolyte Is Dissolved in in 50