The minimum volume of water required to dissolve 0.1 ~g lead (II) chloride to get a saturated solution (K_ SP . of PbCl ₂=3.2 10⁻⁸ ; atomic mass of Pb =207 ~u ) is :

Step-by-step solution

( K _ sp )_ PbCl ₂ =3.2 10⁻⁸=32 10⁻⁹ PbCl ₂ s Pb ²⁺ + 2 s 2 Cl ⁻ K _ sp = [ Pb ²⁺ ] [ Cl ⁻ ]^2 ~K _ sp =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