1. Write the dissolution equilibrium

$PbI_2(s) \rightleftharpoons Pb^{2+} + 2I^-$

So,

$K_{sp} = [Pb^{2+}][I^-]^2$

Given:

$K_{sp} = 8.0 \times 10^{-9}$

2. Consider solubility in $0.1$ M $Pb(NO_3)_2$

Let the solubility of $PbI_2$ in this solution be $s$ mol/L.

Then at equilibrium:

• $[Pb^{2+}] = 0.1 + s$
• $[I^-] = 2s$

Since $s$ will be very small compared to $0.1$,

$[Pb^{2+}] \approx 0.1$

Thus,

$K_{sp} = (0.1)(2s)^2$

3. Substitute the values

$8.0 \times 10^{-9} = 0.1 \cdot 4s^2$

$8.0 \times 10^{-9} = 0.4s^2$

$s^2 = \frac{8.0 \times 10^{-9}}{0.4} = 2.0 \times 10^{-8}$

$s = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4}$

Given $\sqrt{2} = 1.41$,

$s = 1.41 \times 10^{-4}\,\text{mol/L}$

4. Match with the required form

We need:

$s = x \times 10^{-6}\,\text{mol/L}$

Since,

$1.41 \times 10^{-4} = 141 \times 10^{-6}$

So,

$x = 141$

5. Check approximation

Since $s = 1.41 \times 10^{-4} \ll 0.1$, neglecting $s$ in $0.1+s$ is valid.

Therefore, the final answer is:

$\boxed{141}$