Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100 . 15 ° C .…

Let a mole $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ be added, dissociation of lead nitrate will result in:

$\begin{array}{ccccc}\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2& \to & {\mathrm{Pb}}^{2+}& +& 2{\mathrm{NO}}_3^{-}\\ \mathrm{a}& & \mathrm{a}& & 2\mathrm{a}\end{array}$

$\begin{array}{cccc}∆{\mathrm{T}}_{\mathrm{b}}=0.15=& 0.5\left[3\mathrm{a}\right]& \Rightarrow & \mathrm{a} = 0.1\\ & {\mathrm{Pb}}_{\left(\mathrm{aq}\right)}^{2+}& +& 2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\to {\mathrm{PbCl}}_2\left(\mathrm{s}\right)\end{array}$
During reformation of lead nitrate as a precipitate,
                                $\begin{array}{ccc}{\mathrm{Pb}}_{\left(\mathrm{aq}\right)}^{2+}& +& 2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\to {\mathrm{PbCl}}_2\left(\mathrm{s}\right)\end{array}$

              $\begin{array}{ccccc}\mathrm{t}=0& & 0.1& & 0.2\\ \mathrm{t}=\infty & & 0.1& & \left(0.2-2\mathrm{x}\right)\end{array}$ ....(i)

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

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}= {\mathrm{k}}_{\mathrm{f}} \times \mathrm{m} \\ \Rightarrow 0.8=1.8\left[\frac{0.3-3\mathrm{x}+0.2+0.2}{1}\right] \end{gathered}$$

$\Rightarrow \mathrm{x}=\frac{2.3}{27}$

From (i),
$$\begin{gathered} \Rightarrow {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^{2+}\right][2{\mathrm{Cl}}^{-}{]}^2 \\ \Rightarrow \left(0.1-\frac{2.3}{27}\right){\left(0.2-\frac{4.6}{27}\right)}^2=13\times {10}^{-6} \end{gathered}$$