Calculate normality of 2.1%(w/V) $H_{2} {S O}_{4}$ solution.

2.14 N

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4.28 N

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0.428 N

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0.214 N

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Solution

The correct option is B 0.428 N
2.1 % (w/v) $H_{2} S O_{4}$ means 2.1 g of $H_{2} S O_{4}$ is present in 100 ml of solution.
No. of equivalents of $H_{2} S O_{4}$ $= \frac{w e i g h t}{g r a m e q u i v a l e n t o f H_{2} S O_{4}} = \frac{2.1}{49} m o l e s$
Normality $= \frac{e q u i v a l e n t w t .}{v o l u m e o f s o l u t i o n}$
$= \frac{2.1}{49} \times 1000$
= 0.428 N

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