WAEC 2020 Chemistry · Question 36

When 250 cm3 of a saturated solution of CuSO 4 at 30°C was evaporated to dryness, 5.0 g of the salt was…

Question

When 250 cm3 of a saturated solution of CuSO $_4$ at 30°C was evaporated to dryness, 5.0 g of the salt was obtained.. What is the solubility of the salt at 30°C? [CuSO $_4$ = 160]

A.0.031
B.0.125
C.0.640
D.1.560

Answer: B. 0.125

Explanation written with AI and checked against the answer key

Analyze the question: A saturated solution of 250 cm³ gives 5.0 g of CuSO₄ on evaporation. We need the solubility, which in WASSCE chemistry is usually given in mol/dm³ (the options are small numbers, so they are molar values). Concept: Solubility (mol/dm³) = (mass of solute per dm³) ÷ molar mass Solution: Step 1: Find the mass in 1000 cm³ (1 dm³). 250 cm³ contains 5.0 g 1000 cm³ contains (5.0 × 1000) ÷ 250 = 20 g Step 2: Convert to moles using the molar mass of 160 g/mol. Moles = 20 ÷ 160 = 0.125 mol Solubility = 0.125 mol/dm³ Answer: B Tip: When the options are small decimals, the solubility is in mol/dm³. First scale the mass up to 1 dm³ (here ×4), then divide by the molar mass. Check the working with the shortcut: 5 g ÷ 160 = 0.03125 mol in 0.25 dm³, so ×4 = 0.125 mol/dm³. Option A (0.031) is the trap answer for students who forget to scale up to 1 dm³.

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