• # question_answer 13) Calculate the mass of sodium acetate $(C{{H}_{3}}COONa)$ required to make 500 mL of 0.315 M solution. Given that the molar mass of sodium acetate is 82.

Molarity of a solution may be calculated as, $M=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)$ Given $M=0.375,V=500mL,{{m}_{B}}=82$ Putting these values in Eqn. (i), we get $0.375=\frac{{{w}_{B}}\times 1000}{82\times 500}$ ${{w}_{B}}=\frac{0.375\times 82\times 500}{1000}=15.375g$