Molarity (M) of an aqueous solution containing x g of anhyd. CuSO₄ in 500 mL solution at 32° C is 2 × 10⁻¹ M. Its molality will be _________ × 10⁻³ m. (nearest integer). [Given density of the solution =1.25 g / mL]

Solution

Chemistry ma’am · 3 steps

Molality needs the mass of water alone, so take the solute's mass out of the solution's mass.

1. Moles of solute. Moles = molarity × volume in litres = 0.2 × 0.5 = 0.1 mol of CuSO₄. The molar mass of CuSO₄ is 63.5 + 32 + 64 = 159.5 g mol⁻¹, so x = 0.1 × 159.5 = 15.95 g.
2. Mass of solvent. Mass of solution = volume × density = 500 × 1.25 = 625 g. Mass of water = 625 - 15.95 = 609.05 g = 0.60905 kg.

   m = nₛₒₗᵤₜₑ/(Wₛₒₗᵥₑₙₜ (kg)) = 0.1/0.60905 = 0.1642 mol kg⁻¹

3. So. Molality = 164.2 × 10⁻³ m, which rounds to 164.

The common slip: 160. Dividing by the solution's mass, 0.625 kg, instead of the water's mass gives 160, not 164.

The idea to keep. To go from molarity to molality, use density to get the mass of the solution, then subtract the solute's mass. Molality always divides by kilograms of solvent, never of solution.

Answer from the NTA final answer key. Solution by PrepWiser.