How many grams of copper(II) sulfate pentahydrate (CuSO4 • 5 H20) 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 0.15 M solution, we first need to calculate the molar mass of CuSO4 • 5 H2O:
Copper (Cu) = 63.55 g/mol
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol
Hydrogen (H) = 1.01 g/mol
Molar mass of CuSO4 • 5 H2O:
Cu: 1 * 63.55 g/mol = 63.55 g/mol
S: 1 * 32.06 g/mol = 32.06 g/mol
O: 4 * 16.00 g/mol = 64.00 g/mol
H: 10 * 1.01 g/mol = 10.10 g/mol
Total molar mass = 63.55 + 32.06 + 64.00 + 10.10 = 169.71 g/mol
Now, we can calculate the number of moles of copper(II) sulfate pentahydrate needed:
0.15 M solution means 0.15 moles of copper(II) sulfate pentahydrate in 1 liter of solution (1000 mL).
Number of moles = Molarity * Volume (L)
Number of moles = 0.15 moles/L * 0.100 L = 0.015 moles
Now, we convert moles to grams using the molar mass:
Mass (g) = Number of moles * Molar mass
Mass (g) = 0.015 moles * 169.71 g/mol = 2.54565 grams
Therefore, 2.54565 grams of copper(II) sulfate pentahydrate are needed to prepare 100.00 milliliters of a 0.15 M copper(II) sulfate solution.