$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

  • A
    $20 \ mL$ of $0.5 \ M$ $C_2H_2O_4$
  • B
    $50 \ mL$ of $0.1 \ M$ $C_2H_2O_4$
  • C
    $50 \ mL$ of $0.5 \ M$ $C_2H_2O_4$
  • D
    $20 \ mL$ of $0.1 \ M$ $C_2H_2O_4$

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