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Question

KMnO4 reacts with oxalic acid according to the equation
2MnO4+5C2O24+16H+2Mn2++10CO2+8H2O
Here , 20mL of 0.1 M KMnO4 is equivalent to:

A
120mL of 0.25 M H2C2O4
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B
50mL of 0.10 M H2C2O4
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C
25mL of 0.20 M H2C2O4
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D
50mL of 0.20M H2C2O4
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Solution

The correct options are
B 50mL of 0.10 M H2C2O4
C 25mL of 0.20 M H2C2O4
2MnO4+5C2O24+16H+2Mn2++10CO2+8H2O
20 ml of 0.1 M KMnO4 contains 0.1×0.020 moles of KMnO4 i.e, 0.002 moles of KMnO4
(A) 120 mL of 0.25M H2C2O4 contains 0.25×0.12 moles of H2C2O4 i.e, 0.03 moles of H2C2O4
(B) 50 mL of 0.10 M H2C2O4 contains 0.05×0.1 moles of H2C2O4 i.e, 0.005 moles of H2C2O4
(C) 25 mL of 0.20 M H2C2O4 contains 0.005 moles of H2C2O4
(D) 50 mL of 0.20 M H2C2O4 contains 0.01 moles of H2C2O4
By stiochimetry, 2 moles of KMnO4 is equivalent to 5 moles of H2C2O4
Therefore, 0.002 moles of KMnO4 is equivalent to 0.005 moles of H2C2O4.
Therefore, (B) and (C) are correct options.

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