Question

Calculate the molality, molarity, and mole fraction of FeCl3 in a 21.6 mass % aqueous solution (d = 1.280 g/mL).

molality _______ m

molarity _______M

mole fraction ________

Answer 1

1. Molar mass of FeCl₃ = 162.2 g/mol

Consider 1 kg = 1000 g of the solution

Mass of FeCl₃ = 21.6/100 x 1000 = 216 g

Moles of FeCl₃ = mass of FeCl₃/molar mass of FeCl₃

= 216/162.2 = 1.332 mol

Mass of water = 1000 - 216 = 784 g = 0.784 kg

Molality = moles of FeCl3/mass of water in kg

= 1.332/0.784 = 1.699 mol/kg = 1.7 m

2. Volume of solution = mass of solution/density

= 1000/1.280 = 781.25mL = 0.78125 L

Molarity = moles of FeCl3/volume of solution in L

= 1.332/0.78125 = 1.705 mol/L = 1.705 M

3. Molar mass of water = 18.02 g/mol

Moles of water = mass of water/molar mass of water

= 784/18.02 = 43.51 mol

Mole fraction of FeCl3 = moles of FeCl3/(moles of FeCl3 + moles of water)

= 1.332/(1.332 + 43.51​​​​​​​) = 0.0297