Homework Help: chem- molar weight

Posted by manny on Friday, September 7, 2007 at 7:48pm.

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

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?

chem- molar weight - DrBob222, Friday, September 7, 2007 at 8:03pm
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.
chem- molar weight - manny, Friday, September 7, 2007 at 8:13pm
ok got it!!!!

Thanks for clearing it up for me!

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