A solution contains 10.05 grams of an unknown covalent substance in 50.0

Question

grams of water. The freezing point of this solution is (-) 3.16o C.
What is the formula weight of the unknown compound?

Below is what I tried, but don't know what to do next. Please explain step-by-step. Thanks

m=3.16/1.86=1.69 moles of solute/kg of water. 10.05 x 1000*50 g/kg of water is 200.4 g. So in l liter we have 1.69 moles = 200.4g. Mass of 1 mole =200.4/1.69 = 118.59

Answer 1

I don't follow all of you steps but you have essentially the right answer.
10.05 x 1000*50g/kg is not 200.4 and and kg is 50/1000 and not 50*1000. Anyway, here is the way to do it right.
dT = Kf*m
3.16/1.86 = m = 1.6989 (you threw the last 89 away and you should have rounded up to 1.70).

Tnen m = mols/kg solvent or
m x kg solvent = mols = 1.70 x 0.05 kg = 0.0849 mols.

Then mols = grams/molar mass or
molar mass = grams/mols = 10.05/0.0849 = 118.4.