35. Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution.

(Ksp of PbCl2=3.2×10-8, atomic mass of Pb=207 u)

Suppose, solubility of PbCl2 in water is s mol L-1
PbCl2(s)(1-s)Pb2+(aq)s+2Cl-(aq)2sKsp=[Pb2+].[Cl-]2Ksp=[s][2s]2=4s332×10-8=4s3s3=3.2×10-84=0.8×10-8s3=8.0×10-9
Solubility of PbCl2, s=2×10-3 mol L-1
Solubility of PbCl2 in gL-1=278×2×10-3=0.556 g L-1
(Molar mass of PbCl2=207+(2×35.5)=278)
0.556 g of PbCl2 dissolve in 1 L of water.
0.1 g of PbCl2 will dissolve in=1×0.10.556L of water=0.1798 L
To make a saturated solution, dissolution of 0.1 g PbCl2 in 0.1798 L0.2 L of water will be required.