A solution with 4.17 g of a nonelectrolyte solute in 37 grams of water has a boiling point elevated by 1.20 degree C. What is the molar mass of the solute? (Kb= .51 degree C/m)

Question

_{I understand most of this and I have a grasp on the material I just need a little assistance on this problem. Thank you!}

Answer 1

The formula for boiling point elevation and freezing point depression is...

∆T = imK

∆T = change in boiling point = 1.20ºC

i = van't Hoff factor = 1 for a non electrolyte

m = molality = moles solute/kg solvent = ?

K = boiling point constant = 0.51 ºC/m

Solving for m (molality):

m = ∆T/iK = 1.20ºC/(1)(0.51)

m = 2.35 m = 2.35 mol/kg

Since there are only 37 g of water, that is equal to 0.037 kg, so we can find moles of solute;

2.35 mol/kg x 0.037 kg = 0.0869 moles

molar mass = g/mole = 4.17g/0.0869 mol = 47.99 g/mol = 48 g/mol (to 2 significant figures)