Answer:
- 0.028 °C
Step-by-step explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: ΔTf = Kf.m.
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.