1) How many grams of copper(I1) sulfate pentahydrate (CuSO4 • 5H20) are needelf to prepare 100.00 milliliters of a 0.70 M copper(Il) sulfate solution?

Answer 1

To find the number of grams of copper(II) sulfate pentahydrate needed to prepare 100.00 milliliters of a 0.70 M solution, we can use the formula:

Molarity (M) = moles of solute / liters of solution

First, we need to calculate the moles of copper(II) sulfate needed for the solution. Since the molarity is 0.70 M and the volume is 100.00 milliliters (0.100 L), we can rearrange the formula to solve for moles:

0.70 M = moles / 0.100 L
moles = 0.70 M x 0.100 L
moles = 0.07 mol

The molar mass of copper(II) sulfate pentahydrate (CuSO4 • 5H2O) is approximately 249.7 g/mol. Therefore, the number of grams needed can be calculated as:

grams = moles x molar mass
grams = 0.07 mol x 249.7 g/mol
grams = 17.479 g

Therefore, 17.479 grams of copper(II) sulfate pentahydrate are needed to prepare 100.00 milliliters of a 0.70 M copper(II) sulfate solution.