The mass of oxalic acid crystals $(H_2{C}_2{O}_4.2H_{2}O)$ required to prepare $50$ mL of a $0.2N$ solution is:

The correct option is C $0.63$ g
Molecular weight of oxalic acid $(H_2{C}_2{O}_4.2H_{2}O)$ is $126$ g/mol.

Molarity of $0.2$ N oxalic acid $(H_2{C}_2{O}_4.2H_{2}O)$ solution $=\frac{0.2}{2}=0.1M$

Amount of solution $=50$ mL $=0.05$ L

So, amount of oxalic acid $=126\times 0.1\times 0.05=0.63$ g

Hence, the correct option is $C$