In: Chemistry

# A solution is 3.24 M NH3 (MM 17.04 g/mol) in water (MM 18.02 g/mol). The density...

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?

## Solutions

##### Expert Solution

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).

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