A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm⁻³. The molarity of the solution is ____________ mol dm⁻³. (Round off to the Nearest Integer).
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]

Solution

Chemistry ma’am · 3 steps

Take 1 kg of water, add the KOH mass to get the solution mass, then divide by density for volume.

1. Pick a sample. 6.50 molal means 6.50 mol of KOH in 1000 g of water. The molar mass of KOH is 39 + 16 + 1 = 56 g mol⁻¹, so the solute mass is 6.50 × 56 = 364 g.
2. Solution volume. The total solution mass is 1000 + 364 = 1364 g. Dividing by the density gives V = 1364/1.89 = 721.7 cm³ = 0.7217 dm³.

   M = (6.50 mol)/(0.7217 dm³) = 9.01 mol dm⁻³

3. So. The molarity is about 9.01 mol dm⁻³, which rounds to 9.

The common slip: 12. Dividing only the 1000 g of water by the density, and forgetting the 364 g of KOH, gives a volume of 529 cm³ and a molarity of 12.

The idea to keep. To convert molality to molarity, start from 1 kg of solvent. The solution mass is solvent plus solute, and only that total mass divided by density gives the volume.

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