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equation,
Here, 20 mLof 0.1 M is equivalent to
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a
20 mL of 0.5MH2C2O4
b
50 mL of 0.1MH2C2O4
c
50 mL of 0.01MH2C2O4
d
20 mL of 0.1MH2C2O4
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Correct option is B
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 0.1MH2C2O4Talk to our academic expert!
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