Q.

KMnO4 reacts with oxalic acid according to the equation,2MnO4-+5C2O42-+16H+⟶2Mn2++10CO2+8H2OHere, 20 mLof 0.1 M KMnO4 is equivalent to

Moderate

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By Expert Faculty of Sri Chaitanya

a

20 mL of 0.5MH2C2O4

b

50 mL of 0.1MH2C2O4

c

50 mL of 0.01MH2C2O4

d

20 mL of 0.1MH2C2O4

answer is 2.

(Detailed Solution Below)

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Detailed Solution

2MnO4-+5C2O42-+16H+⟶2Mn2++10CO2+8H2O20 mL of 0.1MKMnO4=20×0.1=2m mol∵ 2 m mol of KMnO4≡5 m mol of C2O42-50 mL of 0.1MH2C2O4=50×0.1=5m mol Hence, 20 mL of 0.1MKMnO4≡50 mL of 0.1MH2C2O4
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