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

4.5 g

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6.3 g

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0.63 g

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0.45 g

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Solution

The correct option is C $0.63 g$
Volume of solution $= 50 m l$

Normality $= 0.2 N$

Eq. wt of oxalic acid $= \frac{126}{2} = 63$

$M a s s = N o r m a l i t y \times E q . w t \times \frac{v o l u m e (m l)}{1000} M a s s = 0.2 \times 63 \times \frac{50}{1000} M a s s = 0.63 g$

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