Verified Solution • 2025–26 Session moderate

QJEEM22Ch081b: JEE Main 2022 Chemistry - Class 11 JEEM22Ch081b Solution: 0.01 M KMnO 4 solution was added to 20.0 mL of 0.05 M Mohr's…

0.01 M KMnO4 solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is _____________ mL. (nearest integer)

Step-by-Step Verified Solution

Meq of oxidizing agent == Meq of reducing agent

(M×V×nF)KMnO4=(M×V×nF)Mohr’s salt \left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\mathrm{KMnO}_{4}}=\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\text {Mohr's salt }}

0.01×20×5=0.05×V×10.01 \times 20 \times 5=0.05 \times V \times 1

Volume required =20 ml=20 ~ \mathrm{ml}

Since initial volume of KMnO4\mathrm{KMnO}_{4} in burette is 50 ml50 ~\mathrm{ml} .

Hence volume of KMnO4\mathrm{KMnO}_{4} left in the burette after end point is 30 ml30 ~\mathrm{ml} .

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