The freezing point of ethanol (C₂H₅OH)is -114.6 °C. The molal freezing point depression constant for ethanol is 2.00 °C/m. What is the freezing point (°C)of a solution prepared by dissolving 50.0 g of glycerin (C₃H₈O₃, a nonelectrolyte)in 200.0 g of ethanol? A)-115.0 B)-5.43 C)-132.3 D)-120.0 E)-114.6

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Quizplus · Accepted Answer

The freezing point depression can be calculated using the formula ΔTf = i * Kf * m. Since glycerin is a nonelectrolyte, i = 1. The molality (m) is calculated as moles of solute per kg of solvent. Moles of glycerin = 50.0 g / 92.09 g/mol = 0.543 mol. Molality = 0.543 mol / 0.200 kg = 2.715 m. ΔTf = 1 * 2.00 °C/m * 2.715 m = 5.43 °C. The new freezing point = -114.6 °C - 5.43 °C = -120.0 °C.

The Freezing Point of Ethanol (C2H5OH)is -114

The Freezing Point of Ethanol (C₂H₅OH)is -114