\( \mathrm{W}=\frac{\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V}(\text { in } \mathrm{mL})}{1000}=\frac{0.75 \times 36.5 \times 25}{1000}\)
\( =0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \)
\( \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH} \)
\(36.5 \mathrm{~g} \mathrm{HCl}\) 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 \(=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}\)
$($ મોલર દળ $Fe=56\, g\, mol^{-1}$, $Cl=35.5\, g\, mol^{-1})$
(આપેલ : $Mg$ નો પરમાણ્વીય દળ $24\, g\, mol ^{-1} ; N _{ A }=6.02 \times 10^{23}\, mol ^{-1}$ )