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In the reaction $\mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3^{--} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{--}$ equivalent weight of iodine will be equal to
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The correct answer is:
$1 / 2$ of molecular weight
Molecular weight/2 = Equivalent weight of lodine.
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