A solution is 12.5% Na3PO, by mass. How many grams of Na3PO4 are there in a 35.9mL solution of Na3PO, if the density of the
solution is 1.10g/mL?

Explanation:

Mass of solution of Na₃PO₄ = volume  × density

                                              =   35.9 mL × 1.10 g/mL

                                              =   39.49 g

Now, since the solution only contains 12.5% Na₃PO₄, then the mass of Na₃PO₄ is 12.5% of 39.49 g

                                 12.5% × 39.49 g = 4.94 g          

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%