# Difference between revisions of "User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/11/13"

.050 L of water × $\displaystyle \frac{0.05 mol}{1L}$ = .0025 mol of Na2HPO4 × $\displaystyle \frac{268.07 g}{1 mol}$ = 0.6702 g