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  4. The equivalent weight of potassium permanganate (KMnO4) in neutral medium will be equal to

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Q. The equivalent weight of potassium permanganate $ (KMn{{O}_{4}}) $ in neutral medium will be equal to

BHUBHU 2009

Solution:

Equivalent weight of $KMnO_4$ in neutral medium is as:
$\overset{+7}{2 KMnO _{4}}+ H _{2} O \rightarrow 2 KOH +\overset{+4}{2 MnO _{2}}+3[ O ] $
$\therefore \text { Equivalent weight }=\frac{\text { molecular weight }}{3}$