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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

30 mL of `0.5MH_2C_2O_4` (oxalic acid )

B

50 mL of `0.1 MH_2C_2O_4` (oxalic acid)

C

20 mL of `0.5M H_2C_2O_4` (oxali acid )

D

10 mL of `0.1 MH_2C_2O_4` (oxalic acid)

Text Solution

Verified by Experts

The correct Answer is:
B
`2 xx "moles of " KMnO_4 = 5 xx ` moles of oxalic acid
mmol of `KMnO_4 = 0.1 xx 20 = 2` mmol
so mmol of `H_2C_2O_4 = (5)/(2) xx 2 =5` mmol.
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