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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}} C\mathrm{C}), while (25.4 \mathrm{ g}^{{}^{\circ}}) of (\mathrm{AY}}_{2}) (molar mass (250\mathrm{ g} \mathrm{ mol}^{-1}))in2kgofwaterconstitutesasolutionwithaboilingpointof(100.0260∘) ) 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 Kkgmol−1\mathrm{ K} \mathrm{ kg} \mathrm{ mol}^{-1}) Which of the following is correct ?

Options

  1. A(\mathrm{AX}_2) is fully ionised while (\mathrm{AY}_2) is completely unionised.
  2. B(\mathrm{AX}_2) is completely unionised while (\mathrm{AY}_2) is fully ionised.
  3. C(\mathrm{AX}_2) and (\mathrm{AY}_2) (both) are completely unionised.
  4. 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 AX2\mathrm{AX}_2ΔTb=i K m0.0156=i×0.52×1.24124×13=i3=1+2α1=α\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 $\mathrm{AY}_2\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{~K}_{\mathrm{b}} \mathrm{~m} \\ & 0.0260=\mathrm{i} \times 0.52 \times \frac{25.4}{250 \times 2} \\ & \mathrm{i} \approx 1 \end{aligned}$∴AX2\therefore A X_2 is completely ionised &AY2\& A Y_2 is completely unionised
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