1. **Stating the problem:** We have two solutions, A and B, each containing 0.1 mol of solute in 100 g of water. Solution A is a weak electrolyte producing 2 ions with 40% ionization, and solution B is a strong electrolyte producing 4 ions. We need to find the difference in boiling point elevation ($\Delta T_b$) between these two solutions. Given: $K_b$ for water = 0.52 °C/m. 2. **Formula used:** Boiling point elevation is calculated by: $$\Delta T_b = i \cdot K_b \cdot m$$ where: - $i$ = van't Hoff factor (number of particles the solute dissociates into, adjusted for ionization) - $K_b$ = ebullioscopic constant - $m$ = molality of the solution (mol solute/kg solvent) 3. **Calculate molality ($m$):** Mass of solvent = 100 g = 0.1 kg Molality $m = \frac{0.1 \text{ mol}}{0.1 \text{ kg}} = 1$ mol/kg 4. **Calculate van't Hoff factor ($i$) for each solution:** - For A (weak electrolyte): Number of ions if fully ionized = 2 Degree of ionization $\alpha = 0.4$ $$i_A = 1 - \alpha + n \cdot \alpha = 1 - 0.4 + 2 \times 0.4 = 1 - 0.4 + 0.8 = 1.4$$ - For B (strong electrolyte): Fully ionized, producing 4 ions: $$i_B = 4$$ 5. **Calculate boiling point elevation for each:** $$\Delta T_{bA} = i_A \cdot K_b \cdot m = 1.4 \times 0.52 \times 1 = 0.728 \degree C$$ $$\Delta T_{bB} = i_B \cdot K_b \cdot m = 4 \times 0.52 \times 1 = 2.08 \degree C$$ 6. **Calculate the difference:** $$\Delta T = \Delta T_{bB} - \Delta T_{bA} = 2.08 - 0.728 = 1.352 \degree C$$ **Final answer:** The difference in boiling point elevation is **1.352 °C**, which corresponds to option D.