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Q)

In alkaline medium,$KMnO_4$ reacts as follows:$2KMnO_4+2KOH\rightarrow 2K_2MnO_4+H_2O+O$.Therefore,the equivalent weight of $KMnO_4$,will be

$\begin{array}{1 1}(a)\;31.6&(b)\;52.7 \\(c)\;7.0&(d)\;158.0\end{array} $

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A)
$2KMnO_4(+7)\rightarrow K_2MnO_4(+6)$
Change in O.N=1
Eq.wt of $KMnO_4=\large\frac{\text{Mol.weight of }KMnO_4}{\text{Change in O.N}}$
$\Rightarrow \large\frac{158}{1}$$=158$
Hence (d) is the correct answer.
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