What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?

Answer 1

molality is moles of solute per kg of solvent

mole fraction is moles of solute or solvent divided by total moles present

Let's assume we have 1 mole total, then

FeCl₃ = 0.15 mol

H2O = 1 - 0.15 mol = 0.85 mol H₂O

mass H₂O = 0.85 mol x 18 g/mol = 15.3 g x 1 kg/1000 g = 0.0153 kg

molality = mole FeCl₃/kg water

molality = 0.15 mol/0.0153 kg

molality = 9.80 m

(note: molar mass FeCl₃ is not needed to solve this problem)