Question

What is the freezing point of an aqueous 2.65 m calcium chloride solution?

Answer 1

dT = i x Kf x m, where dT = temperature lowering, i = van't Hoff factor, Kf = freezing point depression constant, and m = molality. 

For CaCl2, i = 3 because one mol CaCl2 releases 1 mol Ca2+ and 2 mol Cl- for a total of 3 mol of particles. 

dT = 3 x 1.86 C/molal x 2.65 molal = 14.8 C 

The freezing point of the solution: 
0 - 14.8 C = -14.8 C