1) How many grams of copper (Il) sulfate pentahydrate (CuSO4 • 5 H20) are needed to prepare 100.00 milliliters of a 0.15 M copper (il) sulfate solution?
To calculate the grams of copper (II) sulfate pentahydrate needed to prepare 100.00 milliliters of a 0.15 M solution, we first need to determine the molar mass of CuSO4 • 5H2O.
Cu: 1 atom x 63.55 g/mol = 63.55 g/mol
S: 1 atom x 32.07 g/mol = 32.07 g/mol
O: 4 atoms x 16.00 g/mol = 64.00 g/mol
H2O: 5 molecules x 18.02 g/mol = 90.10 g/mol
Total molar mass = 63.55 + 32.07 + 64.00 + 90.10 = 249.72 g/mol
Now, we can calculate the moles of CuSO4 in 100.00 milliliters of a 0.15 M solution:
Molarity (M) = moles / volume (L)
0.15 M = moles / 0.100 L
moles = 0.15 x 0.100 L = 0.015 moles
Finally, we can convert moles to grams using the molar mass:
0.015 moles x 249.72 g/mol = 3.7468 grams
Therefore, you would need 3.75 grams of copper (II) sulfate pentahydrate to prepare 100.00 milliliters of a 0.15 M copper (II) sulfate solution.