chem- molar weight
posted by manny on .
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?
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.
ok got it!!!!
Thanks for clearing it up for me!