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)
Correct Answer: 7
Solution and Explanation
First Calculate the moles using the formula:
Moles = Molarity × Volume in litres
Moles\(= 0.35 × 0.25=0.0875 mol\)
Now Calculate the mass of sodium acetate:
Mass = moles × molar mass
Mass\(= 0.35 × 0.25 × 82.02 = 7.18 g≈ 7 g\)
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