Question

Calculate the molality, molarity, and mole fraction of FeCl3 in a 29.5 mass % aqueous solution (d = 1.280g/mL)

Answer 1

let us assume we had 100 g solution

volume fo solution = mass /density = 100/1.28 = 78.125 ml = 0.078125 L

29.5 % mass fraction means we have 29.5 g FeC3 in 100 g solution

FeCl3 moles = mass of FeCl3 / Molar mass of FeCl3 = 29.5 /162.2 = 0.1819

solvent mass = 100-29.5 = 70.5 g = 0.0705 kg

water moles = mass of water / molar mass of water = 70.5/18 =3.917

Mol fraction of FeCl3 = moles of FeCl3 / total moles of solution = ( 0.1819/0.1819+3.917) = 0.04438

FeCl3 molality = moles of solute / solvent mass in kg = 0.1819/0.0705 = 2.58

Molarity = moles of solute / solution vol in L = 0.1819 /0.078125 = 2.33