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.

Options

A. `50 "mL"`
B. `100" mL"`
C. `180" mL"`
D. `18" mL"`

Answer

C

Explanation

`underset((1-s))(PbCl_(2))(s)iffunderset(s)(Pb^(2+))(aq)+underset(2" s")(2Cl^(-))(aq)`
`K_(sp)=[Pb^(2+)][Cl]^(2)=[s][2s]^(2)=4s^(3).`
`s=3sqrt((3.2xx10^(-8))/(4))=2xx10^(-3)" mol L"`
Solubility of `PbCl_(2)` in
`gL=278xx2xx10^(-3)=0.556" gL"`
`because0.556" gm of "PbCl_(2)" dissolve in 1 L of water"`
`:.0.1" gm of "PbCl_(2)" will dissolve in "(0.1xx1)/(0.556)L" of water = 180 mL"`