What is the expected change in the freezing point of water for a 0.015 m aqueous solution of alcl3?

- 0.028 °C

Explanation:

Where, ΔTf is the depression in water freezing point (ΔTf = ???°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of AlCl₃ (m = 0.015 m).

∴ ΔTf = Kf.m = (1.86 °C/m)(0.015 m) = 0.0279 °C ≅ 0.028 °C.

∵ ΔTf = freezing point of pure water - freezing point of the solution in presence of AlCl₃.

∴  freezing point of the solution in presence of AlCl₃ = freezing point of pure water - ΔTf = 0.0 °C - 0.028 °C = - 0.028 °C.

first one is the correct answer