Q15 / 75 · JEE Main 2025 (29 Jan Shift 1) · Chemistry
1.24 g of \(\mathrm{AX}_2\) (molar mass \(124 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is dissolved in 1 kg of water to form a solution with boiling point of \(100.0156^{\circ} \mathrm{C}\), while \(25.4 \mathrm{~g}^{\circ}\) of \(\mathrm{AY}_2\) (molar mass \(250 \mathrm{~g} \mathrm{~mol}^{-1}\) ) in 2 kg of water constitutes a solution with a boiling point of \(100.0260^{\circ} \mathrm{C}\).
\(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Which of the following is correct ?
- Correct answerA\(\mathrm{AX}_2\) is fully ionised while \(\mathrm{AY}_2\) is completely unionised.
- B\(\mathrm{AX}_2\) is completely unionised while \(\mathrm{AY}_2\) is fully ionised.
- C\(\mathrm{AX}_2\) and \(\mathrm{AY}_2\) (both) are completely unionised.
- D\(\mathrm{AX}_2\) and \(\mathrm{AY}_2\) (both) are fully ionised.
Solution
For $\mathrm{AX}_2$
$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{~K} \mathrm{~m} \\
& 0.0156=\mathrm{i} \times 0.52 \times \frac{1.24}{124 \times 1} \\
& 3=\mathrm{i} \\
& 3=1+2 \alpha \\
& 1=\alpha
\end{aligned}$
For $\mathrm{AY}_2$
$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{~K}_{\mathrm{b}} \mathrm{~m} \\
& 0.0260=\mathrm{i} \times 0.52 \times \frac{25.4}{250 \times 2} \\
& \mathrm{i} \approx 1
\end{aligned}$
$\therefore A X_2$ is completely ionised $\& A Y_2$ is completely unionised
Hence, the correct answer is option: 'A'.