JEE Main 202529 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 29 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Solutions
JEE Main 2025 · 29 January · Morning Shift · Chemistry · Solutions
1.24 g of (AX _ 2 ) (molar mass (124 g mol^ -1 ) ) is dissolved in...
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}} ), while (25.4 \mathrm{ g}^{{}^{\circ}}) of (\mathrm{AY}}_{2}) (molar mass (250\mathrm{ g} \mathrm{ mol}^{-1}\mathrm{C}).
(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}}_{2} \mathrm{O}\right)=0.52 )
Which of the following is correct ?
Options
- A(\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.
Correct answer
A. (\mathrm{AX}_2) is fully ionised while (\mathrm{AY}_2) is completely unionised.
Step-by-step solution
For For $\mathrm{AY}_2\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{~K}_{\mathrm{b}} \mathrm{~m} \\
& 0.0260=\mathrm{i} \times 0.52 \times \frac{25.4}{250 \times 2} \\
& \mathrm{i} \approx 1
\end{aligned}$ is completely ionised is completely unionised