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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

28mL

B

260mL

C

2.6mL

D

26mL

Text Solution

Verified by Experts

The correct Answer is:
A
`(N_1 V_1)_(KMnO_4) =(N_2V_2)_(H_2O_2)`
`N = 0.05 rArr 11.2 Vol ---- 2N`
`?----------0.05 N`
Vol of `H_2O_2= (11.2 xx 0.05)/2`
vol of `O_2 = (11.2 xx 0.05)/2 xx 100mL`
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