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KMnO(4) react with oxalic acid according...

`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt to

A

20 mL of `0.5 M H_(2) C_(2)O_(4)`

B

50 mL of `0.1 M H_(2)C_(2) O_(4)`

C

50 mL of 0.5 `M H_(2)C_(2)O_(4)`

D

20 mL of `0.1 M H_(2)C_(2)O_(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
`KMnO_(4)` Oxalic acid
`(M_(1)V_(1))/(n_(1)) = (M_(2)V_(2))/(n_(2)) , (20 xx 0.1)/(2) = (M_(2)V_(2))/(5) = , M_(2)v_(2) = 5`
for (b) `M_(1)V_(1) = 50 xx 0.1 = 5`
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