Answer to Question #99829 in General Chemistry for Ky

Question

How many grams of lead (iv) sulfate are in 225.0 mL of a 0.6554 M solution?

Answer 1

n - mole Pb(SO₄)₂ , C concentration solution - 0.6554 M  (mole/L), V - volume solution 225mL=0.225L,

n= V*C=0.225L * 0.6554 mole/L = 0,147465 mole == 0.1475 mole

m - weight (mass, g) Pb(SO₄)₂ ; Molar [mass] Pb(SO₄)₂ =399.3 g/mole

m=n * Molar = 0.1475 mole * 399.3 g/mole = 58,89675 g == 58,90 g

Answer: 58,90 grams of lead (iv) sulfate are in 225.0 mL of a 0.6554 M solution