The solubility product of PbI₂ is 8.0 $\times$ 10^$-$9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x $\times$ 10^$-$6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given $\sqrt 2$ = 1.41]

Given, ${[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}$<br/><br/>To calculate solubility of Pbl₂ in 0.1 M solution of Pb(NO₃)₂,<br/><br/>(I) $$\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $$<br/><br/>(II) $Pb{I_2}(s)$ $\rightleftharpoons$ $\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S}$<br/><br/>$\therefore$ [Pb²⁺] = S + 0.1 $\approx$ 0.1<br/><br/>$\because$ S < < 0.1<br/><br/>Now, K_sp = 8 $\times$ 10^$-$9<br/><br/>[Pb²] [I^$-$]² = 8 $\times$ 10^$-$9<br/><br/>0.1 $\times$ (2S)² = 8 $\times$ 10^$-$9<br/><br/>4S² = 8 $\times$ 10^$-$8 $\Rightarrow$ S = 141 $\times$ 10^$-$6 M<br/><br/>$\therefore$ x = 141