I worked this problem, I would like to know if I did it correctly, I think I missed something, I'm not sure.....

Question

When a 20.0 gram sample of an unknown compound is dissolved in 500 grams of benzene, the freezing pointof the resulting solution was observed to be 3.77 degrees C.  If the freezing point of pure benzene is 5.48 degrees C and the Kf fro benzene is 5.12 degrees C/m, calculate the molar mass of the unknown compound.
 
ÄTf = KfM
    Kf for benzene = 5.12 degrees C/m
    the normal freezing point  = 5.480C 
    ÄT = (5.48-3.77) = 1.71
 
1.71 degrees = (5.12 degrees C/m)(20.0g/MW)(5kg of solvent)
MW = 20.48g/mol
 
Is this correct?

DrBob222 · Answer

delta T is correct at 1.71 C.
Then delta T = Kf*m and
m = molality = 1.71/5.12 = 0.334 mols/kg
mols = g/MW
0.334 = 20/MW/0.5 kg or
0.334 = (20/MW)*(1/0.5kg)
MW = ??
I have something like 120 or so. Check my thinking. Check my arithmetic.

manny · Answer

ok got it!!!! Thanks for clearing it up for me!