
Answer: C. The neutralisation: \(\text{Ba(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O}\). Each mole of barium hydroxide needs two moles of nitric acid (Ba(OH)₂ is diprotic).
Moles of \(\text{Ba(OH)}_2 = 0.016\) mol.
Moles of \(\text{HNO}_3\) needed \(= 2 \times 0.016 = 0.032\) mol.
Volume of \(\text{HNO}_3 = \dfrac{n}{c} = \dfrac{0.032}{1.25} = 0.0256\) dm³ = 25.6 cm³ — option C.
Options A and B forget the 2:1 stoichiometry and effectively use a 1:1 ratio (or worse). Option D (39.1 cm³) doesn't correspond to a clean stoichiometric ratio and is a distractor.
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