JEE Main 2020 — Redox Reactions Question with Solution
From: JEE Main 2020 (Online) 5th September Evening Slot
Question
The volume, in mL, of 0.02 M K2Cr2O7 solution
required to react with 0.288 g of ferrous
oxalate in acidic medium is _______.
(Molar mass of Fe = 56 g mol–1)
(Molar mass of Fe = 56 g mol–1)
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Show full solutionCorrect answer: 50
Correct answer
50
Step-by-step explanation
K2Cr2O7 + FeC2O4 Cr+3 + Fe+3 + CO2
nfactor of K2Cr2O7 = 3 2 = 6
nfactor of FeC2O4 = 1 + 2 = 3
m. eq. of K2Cr2O7 = m. eq. of FeC2O4
=
vol = 50 ml
nfactor of K2Cr2O7 = 3 2 = 6
nfactor of FeC2O4 = 1 + 2 = 3
m. eq. of K2Cr2O7 = m. eq. of FeC2O4
=
vol = 50 ml
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This is a previous-year question from JEE Main 2020, covering the Redox Reactions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.