JEE MainChemistryRedox Reactions
A 0.05 M solution of KMnO ₄ reacts completely with an excess of oxalic acid ( H ₂ C ₂ O ₄ ) in an acidic medium. Identify the correct statements : (A) 200 mL of the KMnO ₄ solution oxidises 0.025 moles of oxalic acid. (B) The equivalent weight of KMnO ₄ in this medium is ( Molecular weight 5 ) . (C) 100 mL of the KMnO ₄ solution produces 0.05 moles of CO ₂ . (D) The n-factor of oxalic acid is 1 . Choose the correct a
Options
- A(A), (B) and (C) only
- B(B) and (D) only
- C(A) and (C) only
- D(A) and (B) only
Correct answer
D. (A) and (B) only
Step-by-step solution
The balanced chemical equation for the reaction in acidic medium is: 2 KMnO ₄ + 5 H ₂ C ₂ O ₄ + 3 H ₂ SO ₄ K ₂ SO ₄ + 2 MnSO ₄ + 10 CO ₂ + 8 H ₂ O Statement (A): Moles of KMnO ₄ = 0.05 0.2 = 0.01 mol. From stoichiometry, 2 moles of KMnO ₄ react with 5 moles of H ₂ C ₂ O ₄ . So, 0.01 moles of KMnO ₄ will oxidise 0.01 5 2 = 0.025 moles of oxalic acid. Statement (A) is correct. Statement (B): In acidic medium, Mn changes its oxidation state from +7 in KMnO ₄ to +2 in Mn ²⁺ . Change in oxidation state = 5 . Thus, n-fac