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 40.0 g of glycerin (C₃H₈O₃, a nonelectrolyte)in 200.0 g of ethanol? A)-115.0 B)-4.34 C)-132.3 D)-118.9 E)-114.6

Question

Quizplus · Accepted Answer

First, calculate the molality of the solution:

moles of glycerin = mass/molar mass = 40.0 g/92.09 g/mol = 0.4344 mol
mass of solvent = 200.0 g ethanol
molality = moles of solute/kg of solvent = 0.4344 mol /0.200 kg = 2.17 mol/kg

Next, use the molal freezing point depression constant and molality to calculate the change in freezing point:

ΔTf = Kf × molality = (2.00°C/m)×(2.17 mol/kg) = 4.34°C

Finally, subtract the change in freezing point from the normal freezing point of ethanol to get the new freezing point:

freezing point = normal freezing point - ΔTf = (-114.6°C) - (4.34°C) = -118.9°C

Thus, the freezing point of the solution is -118.9°C.

The Freezing Point of Ethanol (C2H5OH)is -114

The Freezing Point of Ethanol (C₂H₅OH)is -114