What is the percent by mass of an aqueous solution that is 1.5m ammonium sulfate?

1.5 m (NH₄)₂SO₄ means a solution containing 1.5 moles of (NH₄)₂SO₄ as solute and 1 kg of water as solvent.

Molar mass of (NH₄)₂SO₄ = 14.01×2 + 1.01×8 + 32.06 + 16.00×4 = 132.16 g/mol
Mass of (NH₄)₂SO₄ in the solution = 132.16 × 1.5 = 198 g

Mass of water = 1 kg = 1000 g
Mass of the solution = 1000 + 198 = 1198 g

Percent by mass = (198/1198) × 100% = 16.5%

1.5m ammonium sulfate = (1.5 mol (NH4)2SO4) / (1 kg H2O)

(1.5 mol (NH4)2SO4) x (132.1395 g (NH4)2SO4/mol) = 198.2 g (NH4)2SO4

(198.2 g (NH4)2SO4) / (1000 g H2O) = 0.1982 = 20% (NH4)2SO4 by mass