Question

A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δf of the solution? Show all the steps taken to solve this problem. (4 points)

Answer 1

Answer:

- 0.99 °C ≅ - 1.0 °C.

Explanation:

• We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(0.53 m) = - 0.99 °C ≅ - 1.0 °C.