Answer:
19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.
Explanation:
Required strength of the solution = 2% (w/v)
This means that 2 gram of solute in 100 ml of solution.
Mass of [tex]CuSO_4.5H_2O[/tex] = 2 g
Moles of [tex]CuSO_4.5H_2O[/tex] =[tex]\frac{2 g}{249.68 g/mol}=0.008010 mol[/tex]
Volume of the solution = 100 mL = 0.1 L
Molarity of the solution:
[tex]M=\frac{0.008010 mol}{0.1 L}=0.08010 mol/L[/tex]
0.08010 moles of [tex]CuSO_4.5H_2O[/tex] are present 1 l of the solution.
Then mass of 0.08010 moles of [tex]CuSO_4.H_2O[/tex] will be:
0.08010 mol × 249.68 g/mol = 19.9993 g≈ 19.9 g
19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.
