JEE Main 2022ChemistryRedox ReactionsHardNumerical

JEE Main 2022Redox Reactions Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu2+ is found to be _____×10-2M (nearest integer)

Given : 2Cu2++4I-Cu2I2+I2I2 + 2S2O3-2    2I + S4O6-2

Enter your answer

Show full solutionCorrect answer: 4
Correct answer
4

Step-by-step explanation

neq. of I2=neq of Na2 S2O3=20×0.02×10-3

2×nmol of I2=0.4 ×10-3

nmol of I2=0.2 m mol

nmol of Cu+2=0.2×2×10-3

Cu+2=0.4×10-310×10-3=0.04=4×10-2

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Redox Reactions chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.