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In the reaction $P b + P b O_{2} + H_{2} S O_{4} \to P b S O_{4} + H_{2} O$ the equivalent weight of $H_{2} S O_{4}$ is M = M.wt of $H_{2} S O_{4}$

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$\frac{M}{4}$

$\frac{M}{2}$

answer is A.

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Detailed Solution

$E_{H_{2} S O_{4}} = \frac{2 M}{2} = M$

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