Calculate the molality of methanol ?

Question

In a solution prepared by dissolving 75 mL methanol CH3OH (density=0.791 g/mL) in 150 g ethanol.

Answer 1

ethanol is your solvent. Molality is defined as moles solute/kilogram solvent. (75mL methanol)(0.791g/mL) = 59.325 grams methanol

(59.325 grams)(1 mol/32grams methanol) = 1.85390625 moles methanol

Now, you have 150 g ethanol/1000 = 0.15 Kg ethanol

(1.85390625 moles methanol)/(0.15 Kg ethanol) = 12.m (molal) methanol

This is my answer to two sig figs. If you have any problems with sig figs, here was the full answer: 12.359375

Answer 2

Mass CH3OH = 75 mL x 0.791 g/mL = 59.3 g

Moles CH3OH = mass / molar mass = 59.3 g / 32.0424 g/mol = 1.85

m = 1.85 mol / 0.150 Kg =12.3