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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 30 January · Morning Shift · Chemistry · Solutions

The mass of sodium acetate ( CH _ 3 COONa) required to prepare...

The mass of sodium acetate (CH3COONa)\left({CH}_{3}COONa\right) required to prepare 250mLof0.35M250mL of 0 . 35M aqueous solution is _____ g.\left(\right) Molar mass of {CH}_{3}COONaisis82 . 02g{mol}^{- 1}) Round off to the nearest integer.

Correct answer

0

Step-by-step solution

Given, Molarity= 0.35M0 . 35M

Moles = Molarity ×\times Volume in litres

=0.35×0.25= 0 . 35 \times 0 . 25

No. of moles = Mass/ Molar mass 

Mass = moles ×\timesmolar mass

=0.35×0.25×82.02=7.18g= 0 . 35 \times 0 . 25 \times 82 . 02 = 7 . 18g

Ans. 77

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