The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is sucrose.

Question

The freezing point of water H₂O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is sucrose.

If 10.98 grams of sucrose, C₁₂H₂₂O₁₁ (342.3 g/mol), are dissolved in 218.2 grams of water ...

The molality of the solution is _________ m.

The freezing point of the solution is ________°C.

Answer 1

Molality = mols sucrose per kg of water

mols sucrose = 10.98 g x 1 mol / 342.3 g = 0.03208 mols

kg water = 218.2 g x 1 kg / 1000 g = 0.2182 kg

molality = 0.03208 mols / 0.2182 kg = 0.1470 m (4 sig. figs.)

Freezing point depression = ∆T = imK

∆T = ?

i = van't Hoff factor = 1 for sucrose (non electrolyte)

m = molality = 0.1470 mol/kg

K = freezing constant for water = 1.86º/m

Solving for ∆T we have

∆T = (1)(0.1470)(1.86

T = 0.273º

The freezing point of the solution is -0.27ºC (2 sig. figs.)