The mass of sodium acetate CH 3 COONa required to prepare 250 mL of 0 . 35 M aqueous solution is _____ g.…

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.

Solution

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

Asked in: JEE Main 2024 (30 Jan Shift 1)

Practice more Solutions questions on Aicharya