1. Use boiling point elevation formula

For a solute in water, $\Delta T_b = i K_b m$ where:

• $i$ = van't Hoff factor
• $K_b = 0.52\,\text{kg mol}^{-1}$
• $m$ = molality

For salts of type $AX_2$ or $AY_2$:

• if completely unionised: $i=1$
• if fully ionised: $A^{2+} + 2X^-$ or $A^{2+}+2Y^-$, so total $3$ ions, hence $i=3$

2. For $AX_2$

Given:

• mass of solute $=1.24\,$g
• molar mass $=124\,$g mol$^{-1}$
• mass of solvent $=1\,$kg
• boiling point $=100.015^\circ$C

So, $\Delta T_b = 100.015 - 100.000 = 0.015^\circ C$

Moles of $AX_2$: $n = \frac{1.24}{124} = 0.01\,\text{mol}$

Molality: $m = \frac{0.01}{1} = 0.01\,\text{mol kg}^{-1}$

Now, $0.015 = i(0.52)(0.01)$ $i = \frac{0.015}{0.0052} \approx 2.88$

This is very close to $3$, so $AX_2$ is fully ionised.

3. For $AY_2$

Given:

• mass of solute $=25.4\,$g
• molar mass $=250\,$g mol$^{-1}$
• mass of solvent $=2\,$kg
• boiling point $=100.0260^\circ$C

So, $\Delta T_b = 0.0260^\circ C$

Moles of $AY_2$: $n = \frac{25.4}{250} = 0.1016\,\text{mol}$

Molality: $m = \frac{0.1016}{2} = 0.0508\,\text{mol kg}^{-1}$

Now, $0.0260 = i(0.52)(0.0508)$ $i = \frac{0.0260}{0.026416} \approx 0.984$

This is very close to $1$, so $AY_2$ is completely unionised.

4. Match with options

• $AX_2$ fully ionised
• $AY_2$ completely unionised

Therefore, the correct option is: $\boxed{\text{D}}$

5. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.