What volume of 0.1 mol dm⁻³ NaOH is needed to neutralise 0.45 g of anhydrous ethanedioic acid, (COOH)₂ (M = 90 g mol⁻¹)?
Correct answer C. 100 cm³
Explanation
0.45 g is 0.005 mol of a dibasic acid, needing 0.010 mol NaOH. Volume = 0.010 ÷ 0.1 = 0.1 dm³, which is 100 cm³.
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