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The mass of sodium acetate (CH3COONa) required to prepare 250 mL of 0.35 M aqueous solution is g. (Molar mass of CH3COONa is 82.02 g mol-1)
7
Moles = M × V(L) = 0.35 × 0.250 = 0.0875 mol
Mass = moles × molar mass = 0.0875 × 82.02 = 7.18 g ≈ 7 g
Answer: 7
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