Question

A solution is 3.24 M NH₃ (MM 17.04 g/mol) in water (MM 18.02 g/mol). The density of the resulting solution is 0.974 g/mL. What is the mole fraction, mass percent and molality of NH₃ in the solution?

Answer 1

The solution is 3.24 M in NH₃. Consider 1.00 L of the solution.

Moles of NH₃ in 1.00 L solution = (1.00 L)*(3.24 M) = (1.00 L)*(3.24 mol/L) = 3.24 moles.

Molar mass of NH₃ = 17.04 g/mol; therefore, mass of NH₃ corresponding to 3.24 moles = (3.24 mole)*(17.01 g/mol) = 55.2096 g.

Density of the solution is 0.974 g/mL; therefore, mass of 1.00 L solution = (1.00 L)*(1000 mL/1 L)*(0.974 g/mL) = 974.0000 g.

Mass of water in the solution = (mass of solution) – (mass of NH₃) = (974.0000 g) – (55.2096 g) = 918.7904 g = (918.7904 g)*(1 kg/1000 g) = 0.9187904 kg

Moles of water corresponding toe 918.8876 g water = (mass of water)/(molar mass of water) = (918.7904 g)/(18.02 g/mol) = 50.9872 moles.

Mole fraction of NH₃ in the solution = (moles of NH₃)/(total number of moles) = (3.24 moles)/(3.24 moles + 50.9872 moles) = 3.24/54.2272 = 0.0597 ≈ 0.06 (ans).

Mass percent NH₃ = (mass of NH₃)/(mass of solution)*100 = (55.2096 g)/(974.0000 g)*100 = 5.6683% ≈ 5.67% (ans).

Molality of NH₃ = (moles of NH₃)/(kg of water) = (3.24 moles)/(0.9187904 kg) = 3.5263m ≈ 3.53m (ans).