WAEC Chemistry 2013 Objective — Question 20

Question 20 of 50 from the West African Examinations Council (WAEC) Chemistry 2013 Objective paper, with the correct answer and a full explanation.

The volume of 0.25 moldm⁻³ solution of KOH that would yield 6.5g of solid KOH on evaporation is (K = 39.0; O = 16.0; H = 1.00)

A. 464.30 cm³
B. 625.00 cm³Correct
C. 1000.00 cm³
D. 2153.80 cm³

Explanation

Molar mass of KOH = (39+16+1) g/mol = 56 g/mol. Mole of KOH = 6.5g/56g⁄mol = 0.1161 mol. Mole = conc(moldm⁻³)×volume(dm³). 0.1161 mol = 0.25moldm⁻³×volume. Volume = 0.1161mol/0.25moldm⁻³ = 0.4643 dm³ = 464.3 cm³.

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