Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0·375 molar aqueous solution. Molar mass of sodium acetate is 82·0245g mol–¹.

Solution

1000 mL of the solution contain sodium acetate = 0·375 mole,

therefore space space 500 space mL space of space the space solution space would space contains sodium space acetate space equals space fraction numerator 0.375 over denominator 1000 end fraction cross times 500 space mole space space space space space space space space space equals fraction numerator 0.375 over denominator 1000 end fraction mol therefore space space space Mass space of space sodium space acetate space required space space space space space equals space fraction numerator 0.375 over denominator 2 end fraction cross times space molar space mass space of space sodium space acetate space space space space space space equals space fraction numerator 0.375 over denominator 2 end fraction cross times 82.0245 space equals space 15.380 space straight g

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