JEE Main 2024ChemistrySolutionsMediumNumerical

JEE Main 2024 — Solutions Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

The mass of sodium acetate CH3COONa required to prepare 250 mL of 0.35M aqueous solution is _____ g. Molar mass of CH3COONa is 82.02 g mol-1) Round off to the nearest integer.

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Show full solutionCorrect answer: 7
Correct answer
7

Step-by-step explanation

Given, Molarity= 0.35 M

Moles = Molarity × Volume in litres

=0.35×0.25

No. of moles = Mass/ Molar mass 

Mass = moles ×molar mass

=0.35×0.25×82.02=7.18 g

Ans. 7

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About this question

This is a previous-year question from JEE Main 2024, covering the Solutions 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.