Question

An aqueous solution of ammonium sulfate is prepared by dissolving 2.78 g of ammonium sulfate in 8.38×10² g of water. The density of the solution is 1.73 g mL^-1.

a) Determine the mass percent of ammonium sulfate in the solution.
b)Determine the mole fraction of ammonium sulfate in the solution.

Answer 1

Molar mass of ammonium sulphate,(NH₄)₂SO₄ is

                 = (2xAt.mass of N + (8xAt.mass of H) + At.mass of S + (4xAt.mass of O)

                 = (2x14)+(8x1)+32+(4x16)

                 = 132 g/mol

Given mass of (NH₄)₂SO₄ is 2.78 g

Number of moles of (NH₄)₂SO₄ is n = mass/molar mass

                                                   = 2.78 g / 132 (g/mol)

                                                   = 0.021 mol

Mass of water = 8.38x10² g = 838 g

Total mass of the solution = masss of (NH₄)₂SO₄ + mass of water

                                      = 2.78 + 838

                                      = 840.78 g

Molar mass of water, H₂O is = (2xAt.mass of H ) + (At.mass of O )

                                         = ( 2x1) + 16

                                         = 18 g/mol

Number of moles of water , n' = mass/molar mass

                                           = 838 g / 18 (g/mol)

                                           = 46.5 mol

Total number of moles , N = n + n' = 0.021+46.5 = 46.521 mol

(a) mass percent of (NH₄)₂SO₄ is = ( mass of (NH₄)₂SO₄ / mass of solution ) x 100

                                                 = (2.78 / 840.78) x 100

                                                 = 0.33

(b) Mole fraction of (NH₄)₂SO₄ is , X = n / N

                                                    = 0.021 / 46.521

                                                    = 4.51x10^-4