Question

The experimentally measured freezing point of a 1.50 m aqueous solution of CaCl₂ is -6.70°C. The freezing point depression constant for water is K_f = 1.86°C/m. Assume the freezing point of pure water is 0.00°C.

The experimentally measured freezing point of a 1.50 m aqueous solution of CaCl₂ is -6.70°C. The freezing point depression constant for water is K_f = 1.86°C/m. Assume the freezing point of pure water is 0.00°C.

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

____C

Answer 1

i for CaCl2 is 3 has it dissociates into 1 Ca2+ and 2 Cl-
lets now calculate ΔTf
ΔTf = i*Kf*m
= 3*1.86*1.5
= 8.37 oC

This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 8.37
= -8.37 oC
Answer: -8.37 oC