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100 mL of H(2)O(2) is oxidized by 100 mL...

100 mL of `H_(2)O_(2)` is oxidized by 100 mL of `1M KMnO_(4)` in acidic medium (`MnO_(4)^(-)` reduced to `Mn^(+2)` ) 100 mL of same `H_(2)O_(2)` is oxidized by v mL of `1M KMnO_(4)` in basic medium (`MnO_(4)^(-)` reduced to `MnO_(2)`). Find the value of v:

A

500

B

100

C

`100//3`

D

`500//3`

Text Solution

Verified by Experts

The correct Answer is:
D
In acidic medium
Eqts of `H_(2)O_(2)` = Eqts of `KMnO_(4)`
`(100)/(1000)xxMxx2=(100)/(1000)xx1xx5impliesM=2.5`
In basic medium
Eq. of `H_(2)O_(2)` = Eqts of `KMnO_(4)`
`(100)/(1000)xx2.5xx2=(V)/(1000)xx1xx3impliesV=(500)/(3)`
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