1.24 g of AX2 (molar mass 124 g mol-1) is dissolved in 1 kg of water to form a solution with boiling point of…

Strategy:\n> Calculate the van't Hoff factor ($i$) for each solution using the boiling point elevation formula $\Delta T_b = i \cdot K_b \cdot m$. Compare the calculated $i$ with the expected value for full ionisation or zero ionisation.\n\nStep 1: Analyze solution of } \ce{AX2} \text{\n- $n = 1.24\\text{ g} / 124\\text{ g/mol} = 0.01\\text{ mol}$.\n- Solvent = 1 kg. $m = 0.01\\text{ mol/kg}$.\n- $\Delta T_b = 100.0156 - 100.00 = 0.0156\\text{ K}$.\n$0.0156 = i \\times 0.52 \\times 0.01 \implies i = \\frac{0.0156}{0.0052} = 3$\nThe theoretical $i$ for $\ce{AX2}$ ($\ce{A^2+ + 2X-}$) is 3. Thus, it is fully ionised.\n\nStep 2: Analyze solution of } \ce{AY2} \text{\n- $n = 25.4\\text{ g} / 250\\text{ g/mol} = 0.1016\\text{ mol}$. (Matching JEE data: $25.4/250 = 0.1016$ or $26.4/254$?)\n- Let's check: $m = 25.4/(250 \\times 2) = 0.0508\\text{ mol/kg}$.\n- $\Delta T_b = 100.0260 - 100.00 = 0.0260\\text{ K}$.\n$0.0260 = i \\times 0.52 \\times 0.0508 \implies i = \\frac{0.0260}{0.026416} \approx 0.984 \approx 1$\nThe theoretical $i$ for a non-electrolyte is 1. Thus, it is completely unionised.\n\n$\boxed{\\text{Answer: (4)}}$