Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 29 January, Morning Shift — Question 9

1.24 g of AX2\mathrm{AX}_{2} (molar mass 124 g mol−1124 \mathrm{~g} \mathrm{~mol}^{-1} ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156∘C100.0156^{\circ} \mathrm{C}, while 25.4 g of AY2\mathrm{AY}_{2} (molar mass 250 g mol−1250 \mathrm{~g} \mathrm{~mol}^{-1} ) in 2 kg of water constitutes a solution with a boiling point of 100.0260∘C100.0260^{\circ} \mathrm{C}. Kb(H2O)=0.52 K kg mol−1\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?

  1. Option A:

    AX2\mathrm{AX}_{2} and AY2\mathrm{AY}_{2} (both) are completely unionised.

  2. Option B:

    AX2\mathrm{AX}_{2} and AY2\mathrm{AY}_{2} (both) are fully ionised.

  3. Option C:

    AX2\mathrm{AX}_{2} is completely unionised while AY2\mathrm{AY}_{2} is fully ionised.

  4. Option D:

    AX2\mathrm{AX}_{2} is fully ionised while AY2\mathrm{AY}_{2} is completely unionised.

    Correct

Answer: D

Step-by-step solution

For AX2:−ΔTb=Kb×m×i\mathrm{AX}_{2}:-\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \times \mathrm{i} 0.0156=0.52×0.011×iAX20.0156=0.52 \times \frac{0.01}{1} \times \mathrm{i}_{\mathrm{AX}_{2}} ⇒iAX2=3⇒\Rightarrow \mathrm{i}_{\mathrm{AX}_{2}}=3 \Rightarrow complete ionisation For AY2:−ΔTb=Kb×m×i\mathrm{AY}_{2}:-\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \times \mathrm{i} 0.026=0.52×0.0508×iAY20.026=0.52 \times 0.0508 \times \mathrm{i}_{\mathrm{AY}_{2}} ⇒iAY2≃1∴\Rightarrow \mathrm{i}_{\mathrm{AY}_{2}} \simeq 1 \therefore complete unionisation

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Abnormal Colligative Properties - van't Hoff Factor
1.24 g of AX 2 (molar mass 124 g mol -1 ) is dissolved in 1 kg of… | JEE Main 2025 PYQ with Solution · DhiX AI