Final answer:
To prepare 50.00 mL of a 0.0800M CuSO4× 5H2O solution, you would need 0.999 grams of cupric sulfate pentahydrate.
Step-by-step explanation:
To calculate the number of grams of cupric sulfate pentahydrate (CuSO₄.5H₂O) needed, you need to use the formula:
grams = volume (L) x molarity (mol/L) x molar mass (g/mol)
First, convert the given volume from milliliters (mL) to liters (L) by dividing it by 1000:
Volume = 50.00 mL / 1000 mL/L = 0.05000 L
Next, substitute the values into the formula:
grams = 0.05000 L x 0.0800 mol/L x (159.62 g/mol + 5(18.02 g/mol))
Solving this equation:
grams = 0.05000 L x 0.0800 mol/L x (159.62 g/mol + 90.10 g/mol)
grams = 0.05000 L x 0.0800 mol/L x 249.72 g/mol
grams = 0.999 g