Question

Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. kf = 1.86ï°c/m

Answer 1

Freezing point depression is a colligative property.

The formula that rules it is:

ΔT f = Kf * m

Where m is the molality of the solution => m = moles of solute / Kg of solvent

=> m = 0.31 mol / 0.175 kg = 1.771 m

And kf = 1.86 °C/m

=> ΔTf = 1.771 m * 1.86 °C / m = 3.3 °C

=> Freezing point of the solution = normal freezing point of water - 3.3 °C = 0 - 3.3°C = - 3.3°C.

Answer: -3.3 °C