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

0.180 M

Molarity=$ \frac{Number\,of\,moles\,of\,MgNH_{4}PO_{4}}{Volume\, of\, solution\,in\,L}=\frac{\frac{Mass\,of\,MgNH_{4}PO_{4}}{Molar\,mass}}{Volume\,of\, solution\,in\,L}$ $=\frac{\frac{6.18\,g}{137.3\,g/mol}}{0.250\,L}=0.180\,M$