Question

In the reaction of $KMnO_4$ with an oxalate in acidic medium, - $MnO^-_4$ is reduced to $Mn^{2+}$ and $C_2O^{2-}_4$ is oxidised to $CO_2$. Hence, $50 \,mL$ of $0.02 \,M\,KMnO_4$ is equivalent to

• A. 100 mL of 0.05 M $H_2C_2O_4$
• B. 50 mL of 0.05 M $H_2C_2O_4$
• C. 25 mL of 0.02 M $H_2C_2O_4$
• D. 50 mL of 0.10 M $H_2C_2O_4$

Answer 1

Answer: (B)

Correct answer is (b) 50 mL of 0.05 M $H_2C_2O_4$