How many grams of copper(II) sulfate pentahydrate (CuSO4 • 5H20) are needed to prepare 100.00 milliliters of a 0.70 M copper(Il) sulfate solution?

Answer 1

To calculate the grams of copper(II) sulfate pentahydrate needed, we first need to calculate 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
H: 10 atoms x 1.01 g/mol = 10.10 g/mol
Total molar mass = 63.55 + 32.07 + 64.00 + 10.10 = 169.72 g/mol

Next, we need to calculate the moles of copper(II) sulfate required for 100.00 ml of 0.70 M solution:

Moles = Volume (liters) x Molarity = 0.1 L x 0.70 mol/L = 0.07 mol

Now, we can calculate the grams required:

Grams = Moles x Molar mass = 0.07 mol x 169.72 g/mol ≈ 11.88 grams

Therefore, 11.88 grams of copper(II) sulfate pentahydrate are needed to prepare 100.00 milliliters of a 0.70 M copper(II) sulfate solution.