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.  

Answer:

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 \[0.375=\frac{{{w}_{B}}\times 1000}{82\times 500}\] \[{{w}_{B}}=\frac{0.375\times 82\times 500}{1000}=15.375g\]