1.24 mathrm~g$1.24 \mathrm{~g}$ of mathrmAX_2$\mathrm{AX}_{2}$ (molar mass 124 mathrm~g mathrm~mol^-1$124 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in 1 mathrm~kg$1 \mathrm{~kg}$ of water to form a solution with boiling point of 100.0156^circ mathrmC$100.0156^{\circ} \mathrm{C}$ , while 25.4 mathrm~g$25.4 \mathrm{~g}$ of mathrmAY_2$\mathrm{AY}_{2}$ (molar mass 250 mathrm~g mathrm~mol^-1$250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in 2 mathrm~kg$2 \mathrm{~kg}$ of water constitutes a solution with a boiling point of 100.0260^circ mathrmC$100.0260^{\circ} \mathrm{C}$ . mathrmK_b(H_2O) = 0.52 \, K \, kg \, mol^-1$\mathrm{K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}}$ Which of the following is correct?

### Formulas Used Elevation in boiling point formula involving the van't Hoff factor (i$i$): Delta T_b = i cdot K_b cdot m$$\Delta T_b = i \cdot K_b \cdot m$$ Where: * Delta T_b = T_b - T_b^circ$\Delta T_b = T_b - T_b^\circ$ (Boiling point elevation) * K_b = textEbullioscopic constant$K_b = \text{Ebullioscopic constant}$ * m = textMolality of the solution left(fractextmoles of solutetextmass of solvent in kgright)$m = \text{Molality of the solution } \left(\frac{\text{moles of solute}}{\text{mass of solvent in kg}}\right)$ ### Core Logic **Step 1: Evaluate solution system mathrmAX_2$\mathrm{AX}_2$** Delta T_b = 100.0156^circmathrmC - 100.0000^circmathrmC = 0.0156^circmathrmC$$\Delta T_b = 100.0156^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0156^\circ\mathrm{C}$$ textMolality m_1 = frac1.24text g / 124text g mol^-11text kg = 0.01text mol/kg$$\text{Molality } m_1 = \frac{1.24\text{ g} / 124\text{ g mol}^{-1}}{1\text{ kg}} = 0.01\text{ mol/kg}$$ Using the elevation formula: 0.0156 = i_mathrmAX_2 cdot 0.52 cdot 0.01$$0.0156 = i_{\mathrm{AX}_2} \cdot 0.52 \cdot 0.01$$ i_mathrmAX_2 = frac0.01560.0052 = 3$$i_{\mathrm{AX}_2} = \frac{0.0156}{0.0052} = 3$$ Since theoretical dissociation of mathrmAX_2 rightarrow mathrmA^2+ + 2mathrmX^-$\mathrm{AX}_2 \rightarrow \mathrm{A}^{2+} + 2\mathrm{X}^-$ produces 3$3$ particles, i = 3$i = 3$ implies that **mathrmAX_2$\mathrm{AX}_2$ is fully ionised**. --- **Step 2: Evaluate solution system mathrmAY_2$\mathrm{AY}_2$** Delta T_b = 100.0260^circmathrmC - 100.0000^circmathrmC = 0.0260^circmathrmC$$\Delta T_b = 100.0260^\circ\mathrm{C} - 100.0000^\circ\mathrm{C} = 0.0260^\circ\mathrm{C}$$ textMolality m_2 = frac25.4text g / 250text g mol^-12text kg = 0.0508text mol/kg$$\text{Molality } m_2 = \frac{25.4\text{ g} / 250\text{ g mol}^{-1}}{2\text{ kg}} = 0.0508\text{ mol/kg}$$ Using the elevation formula: 0.0260 = i_mathrmAY_2 cdot 0.52 cdot 0.0508$$0.0260 = i_{\mathrm{AY}_2} \cdot 0.52 \cdot 0.0508$$ i_mathrmAY_2 = frac0.02600.0264 approx 1$$i_{\mathrm{AY}_2} = \frac{0.0260}{0.0264} \approx 1$$ Since i approx 1$i \approx 1$, it behaves as a non-electrolyte, meaning **mathrmAY_2$\mathrm{AY}_2$ is completely unionised**. Thus, **mathrmAX_2$\mathrm{AX}_2$ is fully ionised while mathrmAY_2$\mathrm{AY}_2$ is completely unionised**. ### Pattern Recognition A van't Hoff factor matching the complete stoichiometric ion count (i = 3$i = 3$ for mathrmAX_2$\mathrm{AX}_2$) confirms complete ionisation, whereas a factor near unity (i = 1$i = 1$) indicates no dissociation into separate ions. **Correct Option:** **(D)**