Calculate the volume of water required to dissolve`0.1g` lead (II) chloride to get a saturaed solution `(K_(sp)` of `PbCI_(2) = 3.2 xx 10^(-8)`, atomic mass of `Pb = 207 u)`. Multiply your answer with 10 to get answer.

Explanation

Suppose the solubility of `PbCl_(2)` in water is S mol `L^(-1) `. Then
`{:(PbCl_(2)(s)hArrPb^(2+)(aq)+2Cl^(-)(aq)),(" "S" " 2
S):}`
`K_(sp)=(S)(2S)^(2)=4S^(3) = 3.2 xx 10^(-8) ` (Given)
or `S^(3) = 0.8 xx 10^(-8) = 8xx10^(-9) or S = 2 xx 10^(-3) "mol" L^(-1)`
Molar mass of `PbCl_(2) = 207 + 71 = 278`
`:. ` Solubility of `PbCl_(2) ` dissolve in 1 L of water to make a saturated solution. Hence, water required to dissolve 0.1 g of `PbCl_(2) = (1L)/(0.556g) xx 0.1 g = 0.1798 L = 179.8 mL `.