An aqueous solution of 6.3g oxalic acid dihydrate is made up to 250 mL, the volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

Equivalents of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_{4}2{\mathrm{H}}_2\mathrm{O}=\mathrm{Equivalents} \mathrm{of} \mathrm{NaOH} (\mathrm{At} \mathrm{equivalence} \mathrm{point})$Strength of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_{4}2{\mathrm{H}}_2\mathrm{O}(\mathrm{in} \mathrm{g}/\mathrm{L})=\frac{6.3}{250/1000}=25.2 \mathrm{g}/\mathrm{L}$Normality of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_{4}2{\mathrm{H}}_2\mathrm{O}=\frac{\mathrm{Strength}}{\mathrm{Eq}.\mathrm{wt}}=\frac{25.2}{63}=0.4 \mathrm{N}$$\left\{\mathrm{Eq}.\mathrm{wt}.\mathrm{of} \mathrm{oxalica} \mathrm{acid}=\frac{\mathrm{Mol}.\mathrm{wt}}{2}=\frac{126}{2}=63\right\} \mathrm{Using} \mathrm{normality} \mathrm{equation}:$$\stackrel{N_1{V}_1=N_2{V}_2}{\left(H_2{C}_2{O}_{4}2H_{2}O\right)\left(NaOH\right)}$$0.4\times 10=0.1\times {\mathrm{V}}_2{\mathrm{orV}}_2=\frac{0.4\times 10}{0.1}=40 \mathrm{mL}$