The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( K s p of P b C l 2 = 3.2 × 10 - 8 ;atomic mass of P b = 207 u ) is:

Step-by-step solution

( K s p ) P b C l 2 = 32 × 10 - 9 PbCl 2 = Pb 2 +     s      + 2 Cl -   2 s K s p =   P b 2 +   2 C l - 2 K s p = 4 s 3   =   32 × 10 - 9 s 3   =   8 × 10 - 9 s = 2 × 10 - 3 M w M . w × 1 V L = 2 × 10 - 3 0.1 278 × 1 V L = 2 × 10 - 3 V L = 0.1 × 1000 278 × 2 = 0.18   L