In acidic medium, 25 mL of 0.1 M H2C2O4 is titrated with 0.02 M KMnO4 (reduced to Mn^2+). The volume of KMnO4 solution required at the end point is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx122

In acidic medium, $25 mL$ of $0.1 M$ $H X_{2} C X_{2} O X_{4}$ is titrated with $0.02 M$ $KMnO X_{4}$ (reduced to $Mn X^{2 +}$ ). The volume of $KMnO X_{4}$ solution required at the end point is:

Milliequivalents of oxalic acid

= 25 \times 0.1 \times 2 = 5

. For

KMnO X_{4}

N = 0.02 \times 5 = 0.1

V = 5/0.1 = 50 mL