1. Find moles of dissolved lead nitrate from boiling point elevation

For lead nitrate, $\text{Pb(NO}_3\text{)}_2 \rightarrow \text{Pb}^{2+} + 2\text{NO}_3^-$ So, assuming complete dissociation, $i = 3$

Boiling point elevation is: $\Delta T_b = iK_b m$ Given: $\Delta T_b = 100.15 - 100 = 0.15^\circ C,$ $K_b = 0.5$

Thus, $0.15 = 3 \times 0.5 \times m$ $m = \frac{0.15}{1.5} = 0.1$

Since 1 litre water is taken and molality is to be taken equal to molarity, $[\text{Pb(NO}_3\text{)}_2] = 0.1\,\text{M}$ Hence initial moles of lead nitrate dissolved: $0.1\,\text{mol}$

So initially:

• $\text{Pb}^{2+} = 0.1\,\text{mol}$
• $\text{NO}_3^- = 0.2\,\text{mol}$

2. Add 0.2 mol NaCl and use freezing point depression

NaCl dissociates as: $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$ So added particles before any reaction:

• $\text{Na}^+ = 0.2\,\text{mol}$
• $\text{Cl}^- = 0.2\,\text{mol}$

Now chloride reacts with lead: $\text{Pb}^{2+} + 2\text{Cl}^- \rightleftharpoons \text{PbCl}_2(s)$

Let final dissolved moles of $\text{Pb}^{2+}$ be $x$. Then moles of $\text{Cl}^-$ left in solution will be: $0.2 - 2(0.1 - x) = 2x$

Also ions remaining in solution are:

• $\text{Pb}^{2+} = x$
• $\text{Cl}^- = 2x$
• $\text{Na}^+ = 0.2$
• $\text{NO}_3^- = 0.2$

Total solute particles in solution after precipitation: $x + 2x + 0.2 + 0.2 = 3x + 0.4$

3. Use freezing point depression data

Given freezing point is $-0.8^\circ C$, so $\Delta T_f = 0.8^\circ C$

Using: $\Delta T_f = K_f \times (\text{total particle molality})$ $0.8 = 1.8(3x + 0.4)$ $3x + 0.4 = \frac{0.8}{1.8} = \frac{4}{9}$ $3x = \frac{4}{9} - \frac{2}{5} = \frac{20 - 18}{45} = \frac{2}{45}$ $x = \frac{2}{135} \approx 0.01481$

Thus, $[\text{Pb}^{2+}] = x = 0.01481$ $[\text{Cl}^-] = 2x = 0.02963$

4. Calculate solubility product

$K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$ $= x(2x)^2 = 4x^3$

Now, $K_{sp} = 4\left(\frac{2}{135}\right)^3$ $= 4\cdot \frac{8}{135^3} = \frac{32}{2460375}$ $\approx 1.3006 \times 10^{-5}$

So, $K_{sp} \approx 13.0 \times 10^{-6}$

5. Nearest integer

Required blank is: $\boxed{13}$