Answer:
2.75 grams
Explanation:
First, we can find out how many moles of copper (II) sulfate are needed by multiplying the molarity by the volume of solution.
1 L = 1000 mL
⇒ 100 mL = 0.1 L
↓↓↓
(0.11 mol / L)(0.1 L) = 0.011 mol
Next, we can multiply this by the molar mass of copper (II) sulfate pentahydrate:
molar mass = (63.546 + 32.066 + 4(15.999)) + 5(2(1.008) + 15.999) g/mol
molar mass = 249.683 g/mol
↓↓↓
(0.011 mol)(249.683 g/mol) = 2.75 g
So, approximately 2.75 grams of copper (II) sulfate pentahydrate are needed to prepare 100 milliliters of a 0.11M copper (II) sulfate solution.
