How would your unknowns calculated molar mass be affected if you spilled some of it after weighing, but before you put it in the solvent?

 

Ok, so less solute means less depression, so delta T is smaller.

 

By this equation: deltaT=Kf x m that would mean calculated molality would be less, which would give you less calculated moles, which would give you a higher molar mass since you are still using the same mass value in the molar mass calculation.

 

So that means that a higher molar mass depresses the freezing point less? I thought it was the other way around. Am I missing something here?

It's the number of moles that matter, since it's a colligative property. A higher molar mass, for a given amount of mass, means fewer moles.

Wow that makes so much sense, thank you.