The weight of crystalline Oxalic acid $\left({\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4·2{\mathrm{H}}_2\mathrm{O}\right)$ required to prepare 100 ml of 0.05N solution is

Given that,

Normality $\left(N\right)=0.05 \mathrm{N}$
Volume $\left(V\right)=100\mathrm{ml}=0.1\mathrm{lt}$

equivalent weight of H₂C₂O₄.2H₂O = $\frac{126}{2}$= 63

mass of  H₂C₂O₄.2H₂O = No. of equivalents x eq.wt

= 0.05 x 0.1 x 63 = 0.315