\(1\) gram of sodium hydroxide was treated with \(25 \mathrm{~mL}\) of \(0.75 \mathrm{M} \mathrm{HCl}\) solution, the mass of sodium hydroxide left unreacted is equal to

\( M=\frac{W \times 1000}{M_2 \times V(\text { in } m L)} \)
\( \mathrm{W}=\frac{\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V}(\text { in } \mathrm{mL})}{1000}=\frac{0.75 \times 36.5 \times 25}{1000}\)
\( =0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \)
\( \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH} \)
\(36.5 \mathrm{~g} \mathrm{HCl}\) reacts with \(\mathrm{NaOH}=40 \mathrm{~g}\)
\(0.684 \mathrm{~g} \mathrm{HCl} \text { reacts with } \mathrm{NaOH}=\frac{40}{36.5} \times 0.684\)\(=0.750 \mathrm{~g}\)
Amount of \(\mathrm{NaOH}\) left \(=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}\)\(=250 \mathrm{mg}\)