Posted by James on Thursday, April 3, 2008 at 4:11pm.
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 = (5.12oC/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
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