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The mass of sodium acetate CH 3 COONa required to prepare 250 mL of 0 . 35 M aqueous solution is _____ g. Molar mass of CH 3 COONa is 82 . 02 g mol - 1 ) Round off to the nearest integer.

Correct answer

0

Step-by-step 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

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