Question

What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of copper (ll) sulfate, CuSO4

Answer 1

Explanation:

$Molarity (M)= \frac{mol}{L}$

First, we need to convert 50.0 mL to L.

1 mL = 0.001 L

50.0 mL = 0.05 L

Now, we need to find the number of moles of CuSO4 when given 10.0 grams of the compound.

159.609 grams CuSO4 = 1 mol CuSO4

10.0 grams CuSO4 = 0.0627 mol CuSO4

Now, we can plug in this information into the formula.

$M=\frac{0.0627}{0.05}$

$M=1.254$

So, the molarity of copper (II) sulfate is 1.254 when given a 50.0 mL aqueous solution.