JEE Main 2019ChemistryRedox ReactionsHardMCQ

JEE Main 2019 — Redox Reactions Question with Solution

JEE Main 2019 (12 Jan Shift 1)

Question

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the amount of NaOH in 50 mL of the given sodium hydroxide solution?

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Show full solutionCorrect option: B
Correct answer
B4 g

Step-by-step explanation

millimoles of oxalic acid =50×0.5

milli equivalent of oxalic acid =50×0.5×2

millimoles of NaOH=25×M= milli equivalent of NaOH

∴50×0.5×2=25×M

⇒M=2

M=2 so, 1000 ml NaOH solution contains

=2 mol of NaOH

=2×40=80g NaOH

∴ 50ml NaOH solution contains =801000×50=4 g

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About this question

This is a previous-year question from JEE Main 2019, 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.