At 25C, the concentration of lead(II) sulfate in a saturated aqueous solution is 7.9 x 10^-M. Determine the volume, in liters, of saturated solution of lead(II) sulfate that contains 1.0g of lead(II) sulfate.
(PbSO4) = 7.9E-4M in a saturated solution.
That is 7.9E-4 mols/L.
Convert to g PbSO4 = mols x molar mass = approx 0.240 but you should do it more accurately.
Therefore, a saturated soltuion contains 0.240 g PbSO4 in 1000 mL solution.
0.240g/1000 mL x ?mL = 1.0g.
Solve for ?mL.