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 solution, it was observed that the solution froze at $$-0.8°$$C. The solutibility product of PbCl$$_2$$ formed is ______ $$\times 10^{-6}$$ at 298 K. (Nearest integer)
Given: K$$_b = 0.5$$ K kg mol$$^{-1}$$ and K$$_f = 1.8$$ kg mol$$^{-1}$$. Assume molality to be equal to molarity in all cases.

Given: Solid Pb(NO₃)₂ dissolved in 1 litre of water boils at 100.15 °C. Then 0.2 mol NaCl is added, and the solution freezes at −0.8 °C.

$$K_b = 0.5$$ K kg mol$$^{-1}$$, $$K_f = 1.8$$ K kg mol$$^{-1}$$. Molality ≈ Molarity.

$$\Delta T_b = 100.15 - 100 = 0.15°\text{C}$$

Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ (van't Hoff factor $$i = 3$$)

$$\Delta T_b = i \cdot K_b \cdot m \implies 0.15 = 3 \times 0.5 \times m$$

$$m = 0.1$$ mol/kg (≈ 0.1 M)

Initial moles: Pb²⁺ = 0.1, NO₃⁻ = 0.2

NaCl → Na⁺ + Cl⁻ (complete dissociation)

Initial species: Pb²⁺ = 0.1, NO₃⁻ = 0.2, Na⁺ = 0.2, Cl⁻ = 0.2

Reaction: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Let $$x$$ mol PbCl₂ precipitate. After equilibrium:

Pb²⁺ = $$(0.1 - x)$$, Cl⁻ = $$(0.2 - 2x)$$, Na⁺ = 0.2, NO₃⁻ = 0.2

Total particle molality = $$0.2 + 0.2 + (0.1-x) + (0.2-2x) = 0.7 - 3x$$

$$\Delta T_f = K_f \times (\text{total particle molality})$$ $$0.8 = 1.8 \times (0.7 - 3x)$$ $$0.7 - 3x = \frac{4}{9}$$ $$3x = 0.7 - \frac{4}{9} = \frac{63 - 40}{90} = \frac{23}{90}$$ $$x = \frac{23}{270}$$ $$[\text{Pb}^{2+}] = 0.1 - \frac{23}{270} = \frac{27 - 23}{270} = \frac{4}{270} = \frac{2}{135}$$ $$[\text{Cl}^-] = 0.2 - \frac{46}{270} = \frac{54 - 46}{270} = \frac{8}{270} = \frac{4}{135}$$ $$K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = \frac{2}{135} \times \left(\frac{4}{135}\right)^2 = \frac{2 \times 16}{135^3} = \frac{32}{2460375}$$ $$K_{sp} \approx 1.3 \times 10^{-5} = 13 \times 10^{-6}$$

To the nearest integer: the value is $$\mathbf{13} \times 10^{-6}$$.