What is the equivalent weight of $H_{2} S O_{4}$ in the reaction?
$H_{2} S O_{4} + N a I \to N a_{2} S O_{4} + I_{2} + H_{2} S + H_{2} O$

12.25

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61.25

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None of these

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Solution

The correct option is A 49
Equivalent weight= $\frac{M o l e c u l a r w e i g h t}{n_{f}}$
As $n_{f}$ of $H_{2} S O_{4}$ is $2$ & molecular weight is $98$ .
Equivalent weight= $\frac{98}{2} = 49$ .
$n_{f}$ of $H_{2} S O_{4}$ is $2$ because it is dibasic.

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