Answer:
1) The solubility product of the lead(II) chloride is
.
2) The solubility of the aluminium hydroxide is
.
3)The given statement is false.
Explanation:
1)
Solubility of lead chloride = 

S 2S
The solubility product of the lead(II) chloride = 
![K_{sp}=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)

The solubility product of the lead(II) chloride is
.
2)
Concentration of aluminium nitrate = 0.000010 M
Concentration of aluminum ion =
Solubility of aluminium hydroxide in aluminum nitrate solution = 

S 3S
The solubility product of the aluminium nitrate = 
![K_{sp}=[Al^{3+}][OH^-]^3](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAl%5E%7B3%2B%7D%5D%5BOH%5E-%5D%5E3)


The solubility of the aluminium hydroxide is
.
3.

Mass of NaCl= 3.5 mg = 0.0035 g
1 mg = 0.001 g
Moles of NaCl = 
Volume of the solution = 0.250 L
![[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M](https://tex.z-dn.net/?f=%5BNaCl%5D%3D%5Cfrac%7B6.0%5Ctimes%2010%5E%7B-5%7D%20mol%7D%7B0.250%20L%7D%3D0.00024%20M)
1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.
![[Cl^-]=[NaCl]=0.00024 M](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D%5BNaCl%5D%3D0.00024%20M)
Moles of lead (II) nitrate = n
Volume of the solution = 0.250 L
Molarity lead(II) nitrate = 0.12 M
![n=0.12 M]\times 0.250 L=0.030 mol](https://tex.z-dn.net/?f=n%3D0.12%20M%5D%5Ctimes%200.250%20L%3D0.030%20mol)
1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.
![[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D%5BPb%28NO_2%29_3%5D%3D0.030%20M)

Solubility of lead(II) chloride = 
Ionic product of the lead chloride in solution :
![Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}](https://tex.z-dn.net/?f=Q_i%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2%3D0.030%20M%5Ctimes%20%280.00024%20M%29%5E2%3D1.7%5Ctimes%2010%5E%7B-9%7D)
( no precipitation)
The given statement is false.