Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 47b

0.683 M

Molarity=$ \frac{Number\,of\,moles\,of\,NaCH_{3}COO}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,of\,NaCH_{3}COO}{Molar\,mass\,of\,NaCH_{3}COO}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{16.8\,g}{82.0\,g/mol}}{0.300L}=0.683\,M$