The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous — JEE Mains Chemistry Past Papers Chemistry Question

💡 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