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 A $0.428 N$
$2.1 %$ $H_{2} S O_{4}$ means $2.1 g r a m$ of $H_{2} S O_{4}$ is present in 100 ml of solution.

No of gram equivalent mass of $H_{2} S O_{4}$ $= \frac{w e i g h t}{g r a m e q u i v a l e n t o f H_{2} S O_{4}}$ $= \frac{2.1}{49}$

$N o r m a l i t y = \frac{E q u i v a l e n t s \times 1000}{v o l u m e o f s o l u t i o n}$
$= \frac{\frac{2.1}{49} \times 1000}{100}$

$= 0.428 N$

Hence, the correct option is $C$

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