The 25°C evaporation of a 100 cm ^ 3 solution of K 2 CO 3 to dryness gave 14g of the salt. What is the…

Analyze the question: A 100 cm³ saturated solution of K₂CO₃ contains 14 g of the salt after evaporation. We need the solubility in mol dm⁻³.

Concept: Solubility (mol dm⁻³) = moles of solute ÷ volume of solution in dm³. Moles = mass ÷ molar mass.

Solution:

1. Moles of K₂CO₃ = 14 g ÷ 138 g/mol = 0.1014 mol

2. Volume = 100 cm³ = 0.1 dm³ (since 1000 cm³ = 1 dm³)

3. Solubility = 0.1014 mol ÷ 0.1 dm³ = 1.014 mol dm⁻³ ≈ 1.01 mol dm⁻³

Common mistake: Option B (0.101) comes from forgetting to convert 100 cm³ to 0.1 dm³. It is just the number of moles, not the concentration.

Tip: Always convert cm³ to dm³ by dividing by 1000 before calculating concentration. A quick shortcut is that for 100 cm³, you multiply moles by 10 to get mol dm⁻³.