Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15^° C. When 0.2 mol of NaCl is added to the resulting…

Solid Lead nitrate is dissolved in $\displaystyle 1$ litre of water. The solution was found to boil at $\displaystyle 100.15^{\circ} \mathrm{C}$. When $\displaystyle 0.2$ mol of NaCl is added to the resulting solution, it was observed that the solution froze at $\displaystyle -0.8^{\circ} \mathrm{C}$. The solubility product of $\displaystyle \mathrm{PbCl}_2$ formed is $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-6}$ at $\displaystyle 298$ K . (Nearest integer)Given : $\displaystyle \mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $\displaystyle \mathrm{K}_{\mathrm{f}}=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. Assume molality to be equal to molarity in all cases.