Normality of $0.1 M$ $H_{3} {P O}_{3}$ is:

0.2 N

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0.3 N

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0.033 N

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0.05 N

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Solution

The correct option is C $0.2 N$
$N o r m a l i t y = \frac{number of equivalent of solute}{1 liter of solution}$

$N o r m a l i t y = M o l a r i t y \times No of equilvalent H$

In $H_{3} P O_{3}$ , there is only two Hydrogen are bonded to oxygens.

So, $N o r m a l i t y = 2 \times M o l a r i t y = 2 \times 0.1 N$

$N o r m a l i t y = 0. 2 N$

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