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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2024 · 4 April · Morning Shift · Chemistry · Redox Reactions

Only (2 ~mL) of (KMnO_4) solution of unknown molarity is required...

Only (2 \mathrm{~mL}) of (\mathrm{KMnO}_4) solution of unknown molarity is required to reach the end point of a titration of (20 \mathrm{~mL}) of oxalic acid ((2 \mathrm{M})) in acidic medium. The molarity of (\mathrm{KMnO}_4) solution should be (\qquad) M.

Correct answer

8

Step-by-step solution

 eq. (KMnO4)= eq. (H2C2O4)M×2×5=2×20×2M=8M\begin{aligned} & \text{ eq. }\left(\mathrm{KMnO}_4\right)=\text{ eq. }\left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\right) \\ & \mathrm{M} \times 2 \times 5=2 \times 20 \times 2 \\ & \mathrm{M}=8 \mathrm{M}\end{aligned}
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