The equation of the reaction is
Let \(H _{2} SO _{4}\) be \(A\), and \(NaOH\) be B
\(\eta_{ A }: \eta_{ B }=1: 2\)
\(\eta_{B}=C_{B} \times V_{B}\)
Where \(C _{ B }-\) is the molar concentration of \(H _{2} SO _{4}\)
\(\& V _{ B }\) - volume of \(H _{2} SO _{4}\) in \(dm ^{3}\).
\(\eta_{B}=0.1 M \times 20 \times 10^{-1} dm ^{3}\)
\(=2 \times 10^{-3} mol\)
\(\frac{\eta_{A}}{\eta_{A}}=\frac{1}{2}, \eta_{B}=\frac{\eta_{A}}{2}\)
\(\eta_{A}=\frac{2 \times 10^{-5}}{2} mol\)
\(=10 \times 10^{3} mol\)
Using \(V_{A}=\frac{\eta_{A}}{C_{A}}\)
\(V_{1}=\frac{1.0 \times 10^{-3}}{0.50}\)
\(=2 \times 10^{-3} dm ^{3}=2.0 cm ^{3}\)