The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous — JEE Mains Chemistry Past Papers Chemistry Question
Question
The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous
Answer: .
💡 Solution & Explanation
Moles = Molarity × Volume in litres = 0.35 × 0.25 Mass = moles × molar mass = 0.35 × 0.25 × 82.02 = 7.18 g Ans. 7 $//(1®
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