Question

A solution prepared by dissolving 3.00 grams of ascorbic acid (vitamin C, C6H8O6), in
50.0 grams of acetic acid has a freezing point that is depressed by
T = 1.33 oC below that of
pure acetic acid. What is the value of the molar freezing point-depression constant for acetic acid?

Answer 1

mols ascorbic acid = grams/molar mass
m = mols/kg solvent

Then 1.33 = Kf*m
Substitut for m and solve for Kf.

Answer 2

^ but what about the i?

Answer 3

I assume the problem expects you to use i = 1.