Question

$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO^{-}_{4}+5C_{2}O^{2-}_{4}+16H^{+}\to2Mn^{2+}+10CO_{2}+8H_{2}O$ 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 $0.1 MC_2H_2O_4$
• C. $20 ml $of $0.5 M C_2H_2O_4$
• D. $20 ml $of $0.1 MC_2H_2O_4$

Answer 1

Answer: (B)

Correct answer is (b) $50 ml $of $0.1 MC_2H_2O_4$