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September 1, 2014

September 1, 2014

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

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 8:03pmdelta 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 8:13pmok got it!!!!

Thanks for clearing it up for me!

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