Question

Calculate the freezing point of a 2.1 m aqueous sucrose solution. (assume that kf for water is 1.86∘c/m.)

Answer 1

Answer: - 3.91 °C


Step-by-step explanation:

1) The depression of the freezing point of a solution is a colligative property and its value is:


ΔTf = m × Kf × i


2) i is the Van't Hoof constant ans for non ionic solutes is 1.


3) m is the molality


m = 2.1 m


3) ΔTf = 2.1 m × 1 × 1.86 °C / mol = 3.91°C.


5) Freezing poin (temperature)

Tf = 0° - 3.91 °C = - 3.91°C