[Vistrum]

Well I came across this question while doing Chemistry Homework tonight and I'm not sure if I've got it right as this is the first time I've ever done anything with Freezing Point Depression.

The freezing point of a solution containing 4.12 g of unknown solute in 100 g of camphor is 166.2 C. Camphor has Kf = 37.7 C/m. Tf = 178.4.

1. What is the freezing point depression for this solution?
I said 12.2 C.
Freezing Point Depression = Tf - Tsolution.

2. What is the molality of the solution of unknown solute in camphor?
I said 0.324 mol/kg.
deltaTf = Kf*m
m = deltaTf/Kf = 12.2 C / 37.3 C.

3. What is the molar mass of the unknown solute?
I said 12.7 g/mol.
molality = mol solute / kg solvent.
mol solute = molality * kg solvent.
MM solute = g solute / mol solute.

If I didn't do this correctly, would anyone mind helping me along? I missed a couple of lectures and had to teach this to myself so I'm really not sure on this.
Thank you!

[Vistrum]

Turns out I was wrong on part 3.

I accidentally skipped this step:
mol solute = molality * kg solvent.

That step made the answer 127 instead of 12.7.

Thank you!