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

0.0416 M

Molarity=$ \frac{Number\,of\,moles\,of\,(NH_{4})_{2}SO_{4}}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,of\,(NH_{4})_{2}SO_{4}}{Molar\,mass}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{2.20\,g}{132.14\,g/mol}}{0.400L}=0.0416\,mol/L=0.0416\, M$