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Chemistry Question on Mole concept and Molar Masses

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)

Answer

First Calculate the moles using the formula:
Moles = Molarity × Volume in litres
Moles =0.35×0.25=0.0875mol= 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=7.18g≈7g= 0.35 × 0.25 × 82.02 = 7.18 g≈ 7 g