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 g of ethanol? A) - 5.42 B) - 120.0 C) - 115 D) - 114.6 E) - 132.3

Question

Quizplus · Accepted Answer

The freezing point depression can be calculated using the formula ΔTf = i * Kf * m. Here, i = 1 for nonelectrolytes, Kf = 2.00 °C/m, and m is the molality. Molality (m) is calculated as moles of solute (glycerin) per kg of solvent (ethanol). Moles of glycerin = 50.0 g / 92.09 g/mol = 0.543 moles. Molality = 0.543 moles / 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 = -120.03 °C, approximately -120.0 °C.

The Freezing Point of Ethanol (C2H5OH) Is - 114

The Freezing Point of Ethanol (C₂H₅OH) Is - 114