In alkaline condition,KMnO_4 reacts as follow | vedclass

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(C) The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.In the given reaction,the oxidation

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$158$

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(C) The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.In the given reaction,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+6$ in $K_2MnO_4$.The change in oxidation state (n-factor) is $|7 - 6| = 1$.The molar mass of $KMnO_4$ is $39 + 55 + (4 	imes 16) = 158 \ g/mol$.Therefore,the equivalent weight = $\frac{158}{1} = 158$.

In alkaline condition,$KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
The equivalent weight of $KMnO_4$ is:

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In alkaline condition,$KMnO_4$ reacts as follows:$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$The equivalent weight of $KMnO_4$ is: