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100 mL of 0.01M KMnO4 oxidises 100mL of ...

100 mL of 0.01M `KMnO_4` oxidises 100mL of acidified `H_2O_2` . The volume of Oxygen in `(MnO_4^(-))` changes to `Mn^(2+)` in acidic medium and to `MnO_4` in alkaline medium).

A

26 mL

B

260 mL

C

`2.6 mL`

D

26 Lt

Text Solution

Verified by Experts

The correct Answer is:
A
`(N_(1)V_(1))_(KMnO_(4))=(N_(2)V_(2))_(H_(2)O_(2))`
`N=0.05 rArr 11.2` Vol of `H_(2)O_(2)=(11.2xx0.05)/(2)`
Vol of `O_(2)=(11.2xx0.05)/(2)xx100 mL`
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