1. **State the problem:** Calculate the difference in boiling point elevation between two solutions given their boiling point elevations. 2. **Formula used:** The boiling point elevation is calculated by the formula $$\Delta T_b = i \cdot K_b \cdot m$$ where: - $i$ is the van't Hoff factor, - $K_b$ is the ebullioscopic constant, - $m$ is the molality of the solution. 3. **Given values:** - For solution B: $i_B = 4$, $K_b = 0.52$, $m = 1$ - For solution A: $\Delta T_{bA} = 0.728$ °C (already calculated) 4. **Calculate $\Delta T_{bB}$:** $$\Delta T_{bB} = 4 \times 0.52 \times 1 = 2.08 \text{ °C}$$ 5. **Calculate the difference $\Delta T$:** $$\Delta T = \Delta T_{bB} - \Delta T_{bA} = 2.08 - 0.728 = 1.352 \text{ °C}$$ 6. **Final answer:** The difference in boiling point elevation is **1.352 °C**, which corresponds to option D.