Solid Lead nitrate is dissolved in 1 litre of water. The solution...

Let a mole $Pb{\left({NO}_{3}\right)}_{2}$ be added, dissociation of lead nitrate will result in:

$\begin{cases}Pb{\left({NO}_{3}\right)}_{2} & \to & {Pb}^{2 +} & + & 2{NO}_{3}^{-} \\ a & & a & & 2a\end{cases}$

$\begin{cases}∆ {T}_{b} = 0 . 15 = & 0 . 5\left[3a\right] & ⇒ & a = 0 . 1 \\ & {Pb}_{\left(aq\right)}^{2 +} & + & 2{Cl}_{\left(aq\right)}^{-} \to {PbCl}_{2}\left(s\right)\end{cases}$
During reformation of lead nitrate as a precipitate,
$\begin{cases}{Pb}_{\left(aq\right)}^{2 +} & + & 2{Cl}_{\left(aq\right)}^{-} \to {PbCl}_{2}\left(s\right)\end{cases}$

$\begin{cases}t = 0 & & 0 . 1 & & 0 . 2 \\ t = \infty & & 0 . 1 & & \left(0 . 2 - 2x\right)\end{cases}$ ....(i)

In final solution after addition of 0.2 moles of sodium and chloride ions:

${\Delta T}_{f} = {k}_{f} \times m ⇒ 0 . 8 = 1 . 8\left[\dfrac{0 . 3 - 3x + 0 . 2 + 0 . 2}{1}\right]$

$⇒ x = \dfrac{2 . 3}{27}$

From (i),
$⇒ {K}_{sp} = [ {Pb}^{2 +} ] [ 2{Cl}^{-}{]}^{2} ⇒ \left(0 . 1 - \dfrac{2 . 3}{27}\right){\left(0 . 2 - \dfrac{4 . 6}{27}\right)}^{2} = 13 \times {10}^{- 6}$