Question

A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent. 
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513 
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K