Question

4. A 500 g of ammonium chloride solution is 3 % by mass with a density of 1.024 g/mL. Calculate molarity and molality of a solution.

Answer 1

3% by mass means 3 g NH4Cl in 100 g solution or 15 g NH4Cl in 500 g solution. With a density of 1.024 g/mL that 500 g will have a volume = mass/density = 500/1.024 = 488 mL.

mols NH4Cl in 15.0 g = g/molar mass = 15.0/53.5 = 0.28
So M = mols/L = 0.28 mols/488 mL = ? M
m = molality = mols/kg solvent = 0.28/0.500 = ? m.