Respuesta :
The solubility of lead(II) chloride is 0.45 g/100 mL of solution. The value of [tex]K_{sp}[/tex] of [tex]PbCl_{2}[/tex] is 1.7×[tex]10^{-5}[/tex] .
So, option C is correct one.
Calculation,
Number of moles of [tex]PbCl_{2}[/tex] = mass/molar mass = 0.45/278.1 = 1.618×[tex]10^{-3}[/tex] moles
Concentration of [tex]Pb^{+2}[/tex] and [tex]Cl^{-}[/tex] ion
Molarity of [tex]Pb^{+2}[/tex] = Number of moles of solute /volume in lit
Molarity of [tex]Pb^{+2}[/tex] = 1.618×[tex]10^{-3}[/tex] moles /0.1 lit = 0.0162 M
[tex]PbCl_{2}[/tex] → [tex]Pb^{+2}[/tex] + 2[tex]Cl^{-}[/tex]
For certain amount of [tex]Pb^{+2}[/tex] , twice this amount of [tex]Cl^{-}[/tex] ion formed.
The concentration of [tex]Cl^{-}[/tex] ion formed = 2× 0.0162 M = 0.0324 M
Now,
The solubility of lead(II) chloride of solution = 0.45 g/100 mL
[tex]K_{sp}[/tex] of [tex]PbCl_{2}[/tex] = [ [tex]Pb^{+2}[/tex] ][tex][Cl^{-}]^{2}[/tex] = 0.0162 ×[tex][0.0324 ]^{2}[/tex] = 1.7×[tex]10^{-5}[/tex] .
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