Molarity of $0.1 N$ oxalic acid is :

0.05

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0.1

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0.2

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0.3

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Solution

The correct option is A $0.05$ M

No. of acidic Hydrogen in oxalic acid, $H_{2} C_{2} O_{4}$ $= 2$

We know that :

$N o r m a l i t y = m o l a r i t y \times a c i d i c - H$

$0.1 = M o l a r i t y \times 2$

$M o l a r i t y = \frac{0.1}{2}$

$= 0.05 M$

Therefore, the option is A.

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