JEE Main 2021ChemistryRedox ReactionsMediumNumerical

JEE Main 2021 — Redox Reactions Question with Solution

JEE Main 2021 (20 Jul Shift 1)

Question

250 mL of 0.5M NaOH was added to 500 mL of 1M HCl. The number of unreacted HCl molecules in the solution after the complete reaction is p×1021. Find out p
(Nearest integer) NA=6.022×1023

Enter your answer

Show full solutionCorrect answer: 226
Correct answer
226

Step-by-step explanation

We known that no. of moles =Vlitre × Molarity & No. of millimoles =Vml× Molarity so millimoles of NaOH=250×0.5
=125
Millimoles of HCl=500×1=500
Now reaction is

so millimoles of HCl left =375 Moles of HCl=375×10-3
No. of HCl molecules =6.022×1023×375×10-3
=225.8×1021
≃226×1021=226

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 2021, 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.