What volume of 0.10 mol/dm³ NaOH exactly neutralises 0.30 g of ethanoic acid (M = 60 g/mol)?
Correct answer C. 50 cm³
Explanation
Moles of acid = 0.30/60 = 5.0 × 10⁻³ mol, reacting 1:1 with NaOH. Volume = n/c = 5.0 × 10⁻³/0.10 = 0.050 dm³ = 50 cm³.
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