Question

The experimentally measured freezing point of a 1.05 m aqueous solution of AlCl3 is -6.25°C. The freezing point depression constant for water is Kf = 1.86°C/m. Assume the freezing point of pure water is 0.00°C.

Part 1 What is the value of the van't Hoff factor for this solution?

Part 2 What is the predicted freezing point if there were no ion clustering in the solution?

Answer 1

ANSWER:

Given,

molality, m = 1.05 m

K_f = 1.86 °C/m

freezing point of pure water, T_f^o = 0.00 ^oC

freezing point of aqueous solution of AlCl₃, T_f = -6.25°C

T_f = T_f^o - T_f = 0.00 °C - (-6.25°C) = 6.25 °C

Part 1:

we know that,

T_f = i x K_f x m

where,

i = van't Hoff factor

6.25°C = i x (1.86 °C/m) x 1.05 m

i = 3.2

Hence, the value of the van't Hoff factor for this solution is 3.2 .

Part 2:

if there were no ion clustering in the solution, then vant's hoff factor is equal to 1.

T_f = i x K_f x m

T_f = K_f x m

T_f = (1.86 °C/m) x 1.05 m = 1.95 ^oC

we know that,

T_f = freezing point of pure water - freezing point of aquous AlCl₃ solution

T_f = T_f^o - T_f

T_f = T_f^o - T_f = 0.00 ^oC - 1.95 ^oC = - 1.95 ^oC

freezing point of aquous AlCl₃ solution = -1.95 ^oC

Hence, freezing point of aquous solution of AlCl₃ is -1.95 ^oC, if there were no ion clustering in the solution.