A pure solvent freezes at 43.4?0°C.The freezing temperature is 40.7°C when 17.5 g of CO(NH₂)₂ is dissolved in 1.00* 10² g of the solvent.Calculate the molal freezing point constant of the solvent.

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Quizplus · Accepted Answer

The freezing point depression is calculated using the formula ΔTf = i * Kf * m. Here, ΔTf = 43.4 - 40.7 = 2.7°C, i = 1 for urea, m = moles of solute/kg of solvent. Moles of urea = 17.5 g / 60.06 g/mol = 0.291 moles. Solvent mass = 100 g = 0.1 kg. Molality (m) = 0.291 moles / 0.1 kg = 2.91 m. Solving for Kf gives Kf = ΔTf / m = 2.7°C / 2.91 m = 0.93 °C/m.

A Pure Solvent Freezes at 43