The molar solubility of PbI2 in a solution of Pb(NO3)2 can be determined using the solubility product constant, Ksp. The solubility product constant is an equilibrium constant that represents the extent to which a solid compound dissolves in a solution. In the case of PbI2, the solubility product constant can be expressed as:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

The equilibrium expression for the solubility product constant is:

Ksp = [Pb2+][I-]^2

where [Pb2+] represents the concentration of Pb2+ ions in the solution and [I-] represents the concentration of I- ions in the solution.

To determine the molar solubility of PbI2, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every mole of PbI2 that dissolves, one mole of Pb2+ ions and two moles of I- ions are formed. Therefore, the molar solubility of PbI2 can be represented as:

[PbI2] = [Pb2+] = 2[I-]

Now let's consider the given information that the solution contains 0.2 moles of Pb(NO3)2. Since Pb(NO3)2 dissociates into one mole of Pb2+ ions and two moles of NO3- ions, we can determine the initial concentrations of the ions:

[Pb2+] = 0.2 moles/L
[I-] = 2 × [Pb2+] = 2 × 0.2 moles/L = 0.4 moles/L

Substituting these values into the solubility product expression, we have:

Ksp = [Pb2+][I-]^2
Ksp = (0.2 moles/L)(0.4 moles/L)^2
Ksp = 0.032 moles^3/L^3

To find the molar solubility of PbI2, we can rearrange the equation [PbI2] = [Pb2+] = 2[I-]:

[PbI2] = 2 × 0.4 moles/L = 0.8 moles/L

So, the molar solubility of PbI2 in the given solution is 0.8 moles per liter.

In conclusion, the molar solubility of PbI2 in a solution of Pb(NO3)2 can be determined using the solubility product constant, Ksp. By considering the stoichiometry of the reaction and the initial concentrations of the ions, we can calculate the molar solubility.

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