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0.1L of 0.01 M KMnO4 oxidized 100 mL H2O2 in acidic medium. Volume of same KMnO4 required in alkaline medium to oxidize 0.1L of same H2O2 will be (Mn changes to Mn2+ in acidic medium and MnO2 in alkaline medium)

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i cant solve it and i want its solution . plz help me.
by (2.8k points)
You can know the n-factor.
1st case:- n-factor=5 [Mn(7) to Mn(2)]

Now, this is one of the best concepts of mole concept,
"The EQUIVALENTS(normality×volume) of all reactants and products are EQUAL."


Thereby, (100ml)(0.05N)=(100ml)V

Do the same in the next case but n-factor in 2nd case is 3.

Solve it further by yourself, it will help you to gain confidence.

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