Question:medium

In the reaction of KMnO\(_4\) with an oxalate in acidic medium, MnO\(_4^-\) is reduced to Mn\(^{2+}\) and C\(_2\)O\(_4^{2-}\) is oxidized to CO\(_2\). Hence, 50 mL of 0.02 M KMnO\(_4\) is equivalent to:

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In redox reactions, use stoichiometry to relate the volumes and concentrations of reactants involved.
Updated On: Mar 25, 2026
  • 100 mL of 0.05 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
  • 50 mL of 0.05 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
  • 25 mL of 0.2 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
  • 50 mL of 0.10 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
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The Correct Option is B

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