Question

camphor melts at 179.8 degrees, and it has a particularly high freezing point depression constant.(kf=40 degrees/m). when 0.186g of an organic substance of an unknown molar mass is dissolved in 22.01g of liquid camphor, the freezing point of the mixture is found to be 176.7 degrees. what is the molar mass of the solute?

Answer 1

delta T = Kf*molality
Solve for molality.

molality = moles/kg solvent
Solve for moles.

moles = grams/molar mass
Solve for molar mass.

Answer 2

what i have done is:
179.8-176.7/40 = 3.1/40 = 0.0775mol
then,
0.0775*22.01g = 1.706moles
then,
0.186g/1.705mol = 0.109g/mol
then answer is 110g/mol
where have i gone wrong?

Answer 3

m = moles/kg solvent. kg solvent is 0.0221, you didn't convert to kg but kept it in grams. I found 108.6 which rounds to 109 BUT to two significant figures it is 1.1 x 10^1

Answer 4

so doing it your way you then habe to times it by 1000 to get it back to g/mol.

Answer 5

Absolutely not. The answer I have is the answer to stay that way.
The equation for molality is m = moles/kg solvent. You MUST insert kg solvent and not grams.
Then moles = grams/molar mass or rearrange to molar mass = grams/moles. There is no "times it by 1000) here at all. The answer is molar mass, not molar mass/1000. I did make a typo at the end. The answer is 1.1 x 10^2, not 1.1 x 10^1 so 110 is correct BUT that is too many s.f. Sorry about the typo.