The amount of oxalic acid (Eq. wt. 63) required to prepare 500 ml of its 0.10 N solutions is

Amount of oxlalic acid, $w=?$

Volume of solution ,V$=500\text{ mL}$

Normality of solution $=\text{N}=0.1\text{ N}$

The chemical formula of oxalic acid is${\text{H}}_2{\text{C}}_2{\text{O}}_4·2{\text{H}}_2\text{O}$

So, there are two replaceable H atoms present  in oxalic acid.

Hence, the basicity of oxalic acid is 2.

$\text{GEW of}\text{oxalic acid}=\frac{\text{GMW}}{\text{Basicity}}=\frac{126}{2}=63g$

From normality equation,

         $\text{N=}\frac{\text{w}}{\text{GEW}}\times \frac{100}{\text{V}\text{in}\text{mL}}$

After substitution the known data in the above equation, we will get                      

          $0.1=\frac{w}{63}\times \frac{{\cancel{1000}}^2}{\cancel{500}}$           

          $\Rightarrow w=\frac{0.1\times 63}{2}=3.15\text{g}$