JEE Main 2020 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2020 (Online) 3rd September Evening Slot
Question
The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___________.
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Show full solutionCorrect answer: 10
Correct answer
10
Step-by-step explanation
H3PO2 + NaOH NaH2PO2 + H2O
Using Stoichiometry
= 0.1 × VNaOH
VNaOH = 10 ml
Using Stoichiometry
Moles of H3PO2 reacted
1
=
Moles of NaOH reacted
1
= 0.1 × VNaOH
VNaOH = 10 ml
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This is a previous-year question from JEE Main 2020, covering the Some Basic Concepts Of Chemistry 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.