When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0.320 C. The molar mass of the unknown compound is ____

Question

(The freezing point depression constant for benzene is 5.12 C kg/mol. )

Answer 1

well, you know the freezing point depression is equal to mass/molmass *k, and you know grams, and the constant k.

Answer 2

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)

Answer 3

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)

Answer 4

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.

Answer 5

ah i got 2.08 and im still getting it wrong

Answer 6

Isn't delta T = Kb*m?
Then 0.320 = (5.12^oC/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

Answer 7

thank you