Chemistry · Redox Reactions

JEE Main 2024 — 4 April, Shift 1 — Question 78

Only 2 mL of $\mathrm{KMnO}_{4}$ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid ( 2 M ) in acidic medium. The molarity of $\mathrm{KMnO}_{4}$ solution should be $\qquad$ M.

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

eq. $\left(\mathrm{KMnO}_{4}\right)=$ eq. $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)$ $M \times 2 \times 5=2 \times 20 \times 2$ $M=8 M$

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Chemistry
Chapter: Redox Reactions
Topic: n-Factor, Redox Titrations, Self Indicator & Miscellaneous Cases