On evaporation to dryness, 250 cm3 of saturated solution of salt X with relative molar mass 101 gave 50.5g of the. What is the soluility of the salt?

Analyze the question: A 250 cm³ saturated solution of salt X (molar mass 101 g/mol) leaves 50.5 g of solid when evaporated to dryness. We need the solubility in mol dm⁻³.

Concept: Solubility in mol dm⁻³ is the number of moles of solute in 1 dm³ of saturated solution.

moles = mass ÷ molar mass

Solution:

1. Moles of salt in 250 cm³: n = 50.5 ÷ 101 = 0.5 mol

2. Convert the volume to dm³: 250 cm³ = 0.25 dm³

3. Solubility (concentration): = 0.5 mol ÷ 0.25 dm³ = 2.0 mol dm⁻³

Check using mass concentration:

• 50.5 g in 250 cm³ gives 202 g in 1000 cm³.
• 202 ÷ 101 = 2.0 mol dm⁻³ ✔

Answer: B. 2.0 mol dm⁻³

Tip: For solubility questions, follow this routine: find moles (mass ÷ molar mass), then scale to 1 dm³ (multiply by 1000 ÷ volume in cm³). Here 250 cm³ is a quarter of a litre, so multiply the moles by 4: 0.5 × 4 = 2.0.