1) How many grams of copperIl) sulfate pentahydrate (CuS, • 5 H,0) are needed to prepare 100.00 milliliters of a 0.15 M copper(Il) sulfate solution?
To calculate the number of grams of copper(II) sulfate pentahydrate needed to prepare 100.00 milliliters of a 0.15 M solution, we first need to calculate the molar mass of CuSO4 • 5H2O.
Molar mass of CuSO4: 63.55 + 32.07 + (4*16.00) = 159.61 g/mol
Molar mass of H2O: 2*(1.01) + 16.00 = 18.02 g/mol
Total molar mass of CuSO4 • 5H2O: 159.61 + (5 * 18.02) = 249.71 g/mol
Now, we can calculate the number of moles of CuSO4 • 5H2O needed:
0.15 moles/L * 0.100 L = 0.015 moles
Finally, we can calculate the mass of CuSO4 • 5H2O needed:
0.015 mol * 249.71 g/mol = 3.7457 grams
Therefore, 3.75 grams of copper(II) sulfate pentahydrate are needed to prepare 100.00 milliliters of a 0.15 M copper(II) sulfate solution.