The volume of water needed to dissolve 0.0724 grams of lead sulfate is ______ L. Assume no volume change upon addition of the solid.

Answer 1

I looked up the solubility of Pb(SO4) in water and found it to be 0.0404 g / 100 mL.

Since we are looking for the number of liters to completely dissolve it, we can write that

100 mL / 0.0404 g x 0.0724 g = 179.2 mL or 0.179 L.