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 gm of $H_{2} S O_{4}$ is present in 100 ml solution.
$N o r m a l i t y = \frac{No. of equivalents}{V o l . o f s o l u t i o n (i n m l)} \times 1000 = \frac{(2.1 / 98) \times 2}{100} \times 1000 = 0.428$
Option C is correct.

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