You use the Kf for the solvent that is in the problem. This problem wants to know the amount of fructose that can be added to PURE WATER (so you use Kf = 1.86). If benzene acted as the solvent you would use the Kf for benzene, for naphthalene you would use Kf for naphthalene, etc.
If that is your only problem with this one then it should be easy to continue.
delta T = Kf*m
You know delta T is 15.0, Kf is 1.86, solve for m = about 8.06m
m = mols/kg solvent
m = 8.06; solvent = 3 kg.
mols = 8.06m x 3 kg = 24.19
mols = g/molar mass
mols = 24.19; molar mass = 180
g = mols x molar mass = 180 x 24.19 = 4,354 g fructose which I would write as 4.35E3 g. Since the problem wants the 3 kg water to freeze, technically, we would use slightly less than that amount.
At -15.0 C, what is the maximum mass of fructose (C6H12O6) you can add to 3.00 kg of pure water and still have the solution freeze? Assume that fructose is a molecular solid and does not ionize when it dissolves in water.
I'm confused on how you pick the Kf value.. Usually I just use 1.86 for water but I don't understand why. For example, the normal freezing point for benzene is 5.49 so can't I use the Kf value for benzene too? How do you know which Kf value to choose for sure?
Please help. Thanks!
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