chemistry
posted by James .
When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0.320 C. The molar mass of the unknown compound is ____
(The freezing point depression constant for benzene is 5.12 C kg/mol. )

well, you know the freezing point depression is equal to mass/molmass *k, and you know grams, and the constant k.

would the answer be .305? i'm still getting this wrong.
i set up the equation to be
x = .500 g/ (.320 C) (5.12 C) 
would the answer be .305? i'm still getting this wrong.
i set up the equation to be
x = .500 g/ (.320 C) (5.12 C) 
I erred in my molality formula
5.12C= .5(mmass*kgbenzene) * .320
mmass= .5*.320/(5.12*.015)
I get about 50 for the molmass. 
ah i got 2.08 and im still getting it wrong

Isn't delta T = Kb*m?
Then 0.320 = (5.12^{o}C/m)*m
m = 0.320/5.12 = 0.0625 m
molality = mols/kg so
mols = molality*kg = 0.0625 mol/kg*0.015 kg = 9.375 x 10^4 mols.
mols = g/molar mass
molar mass = g/mols = 0.500 g/9.375 x 10^4 = 533.3 
thank you