$40 \ mL$ of $0.1 \ N$ $KMnO_4$ is equivalent to $30 \ mL$ of $KHC_2O_4$ solution. How many $mL$ of $0.1 \ N$ $KOH$ are required to titrate $60 \ mL$ of the same $KHC_2O_4$ solution?

  • A
    $80$
  • B
    $30$
  • C
    $28.57$
  • D
    $35.5$

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