50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is :

Given, 50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O} \\ {\left(\frac{\mathrm{VM}}{\mathrm{n}}\right)}_{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}={\left(\frac{\mathrm{VM}}{\mathrm{n}}\right)}_{\mathrm{NaOH}} \\ \frac{50\times 0.5}{1}=\frac{25\times {\mathrm{M}}_{\mathrm{NaOH}}}{2} \\ {\mathrm{M}}_{\mathrm{NaOH}}=2\mathrm{M} \\ \mathrm{given}, \mathrm{Vol}. \mathrm{of} \mathrm{NaOH}=50\mathrm{ml} \\ \mathrm{M}=\frac{\mathrm{W}}{\mathrm{mol}.\mathrm{wt}.}\times \frac{1000}{\mathrm{V}\left(\mathrm{ml}\right)} \\ 2=\frac{\mathrm{W}}{40}\times \frac{1000}{50} \\ \mathrm{Wt}. \mathrm{of} \mathrm{NaOH}=\frac{2\times 40\times 50}{1000}=4\mathrm{g} \end{gathered}$$