Asked by Ace

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?
Answered by DrBob222
mols ascorbic acid = grams/molar mass
m = mols/kg solvent

Then 1.33 = Kf*m
Substitut for m and solve for Kf.
Answered by Ace
^ but what about the i?
Answered by DrBob222
I assume the problem expects you to use i = 1.
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