b
millimole of \(NaOH =0.24 \times 25\)

\(\therefore \quad \text { millimole of acid }=0.24 \times 25\)

\(\Rightarrow \quad \text { mass of acid }=0.24 \times 25 \times 24.2\,mg\)

\(\text { for pure acid, }\)

\(\qquad V =\frac{ w }{ d } ;( d =1.21\,kg / L =1.21\,g / ml )\)

\(\therefore V =\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3}\)

\(=120 \times 10^{-3}\,ml\)

\(=12 \times 10^{-2}\,ml\)