3.01×10 to the 23rd molecules H2O

The factor is (1 mol/6.02 x 10^23 anythings). 3.01 x 10^23 molecules H2O x (1 mole/6.02 x 10^23 molecules H2O)= 0.500 mole H2O.Mar 16, 2009

Explanation:

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%