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What is the equivalent weight of KMNO_(4) in: Neutral medium

Answer»

Solution :EQUIVALENT weight (MASS) (E) of an OXIDISING agent LIKE is given by, Molecular weight of `KMNO_(4)`
`E=("Molecular weight of "KMnO_(4))/(" Change in the oxidation STATE")`
In neutral medium: The oxidation state of Mn in `KMNO_(4)` changes from +7 to +4.
`therefore E_(KMnO_(4))=(158)/(7-4)=52.67`


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