The mass of oxalic acid crystals $(H_{2} C_{2} O_{4} . 2 H_{2} O)$ required to prepare $50$ mL of a $0.2 N$ solution is:

4.5

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6.3

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0.63

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0.45

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Solution

The correct option is C $0.63$ g
Molecular weight of oxalic acid $(H_{2} C_{2} O_{4} . 2 H_{2} O)$ is $126$ g/mol.

Molarity of $0.2$ N oxalic acid $(H_{2} C_{2} O_{4} . 2 H_{2} O)$ solution $= \frac{0.2}{2} = 0.1 M$

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 $C$

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