1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

$$\begin{aligned} & \mathrm{M}=\frac{\mathrm{W} \times 1000}{\left.\mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)} \\ & \mathrm{W}=\frac{\left.\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)}{1000}=\frac{0.75 \times 36.5 \times 25}{1000} \\ &=0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \\ & \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH} \end{aligned}$$​M=M2​×V (in mL)W×1000​W=1000M×M2​×V (in mL)​=10000.75×36.5×25​=0.684 g (Mass of HCl)36.5 gHCl​+40 gNaOH​⟶HCl+NaOH​ 36.5 g HCl reacts with $\mathrm{NaOH}=40 \mathrm{~g}$NaOH=40 g 0.684 g HCl reacts with $\mathrm{NaOH}=\frac{40}{36.5} \times 0.684 \simeq 0.750 \mathrm{~g}$NaOH=36.540​×0.684≃0.750 g Amount of NaOH left $=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}$=1 g−0.750 g=0.250 g=250mg