In alkaline condition KMnO_4 reacts as follow | vedclass

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(D) The equivalent weight of an oxidizing agent is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$In the given re

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$158.0$

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(D) The equivalent weight of an oxidizing agent is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$In the given reaction,the reduction half-reaction is:$MnO_4^- + e^- 	o MnO_4^{2-}$The change in oxidation state of $Mn$ is from $+7$ to $+6$,which involves the gain of $1$ electron per $KMnO_4$ molecule.Thus,the n-factor is $1$.The molar mass of $KMnO_4 = 39.1 + 54.9 + 4 	imes 16 = 158 \ g/mol$.$	ext{Equivalent weight} = \frac{158}{1} = 158$.

In alkaline condition $KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:

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In alkaline condition $KMnO_4$ reacts as follows:$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$Therefore,its equivalent weight will be: