FeCl₃ has a van't Hoff factor of 3.400. What is the freezing point (in °C) of an aqueous solution made with 0.6100 m FeCl₃? (Kf for water is 1.860 °C/m)

Question

FeCl₃ has a van't Hoff factor of 3.400. What is the freezing point (in °C) of an aqueous solution made with 0.6100 m FeCl₃? (Kf for water is 1.860 °C/m)

Answer 1

Colligative properties:

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3.400

m = molality = 0.6100 m

K = freezing constant = 1.860º/m

∆T = (3.400)(0.6100)(1.860)

∆T = 3.858º

Freezing point = 0º - 3.858º = -3.858ºC