Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 21 January, Morning Shift — Question 55

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2\mathrm{PQ}_{2}. When 1 g of PQ is dissolved in 50 g of solvent ' A '. ΔTb\Delta \mathrm{T}_{\mathrm{b}} was 1.176 K while when 1 g of PQ2\mathrm{PQ}_{2} is dissolved in 50 g of solvent ' A ', ΔTb\Delta \mathrm{T}_{\mathrm{b}} was 0.689 K . ( Kb\mathrm{K}_{\mathrm{b}} of ' A ' =5 Kkgmol−1=5 \mathrm{~K} \mathrm{kg} \mathrm{mol}^{-1} ). The molar masses of elements P and Q (in gmol−1\mathrm{g} \mathrm{mol}^{-1} ) respectively, are :

  1. Option A:

    70110

  2. Option B:

    65145

  3. Option C:

    60,25

  4. Option D:

    25,60

    Correct

Answer: D

Step-by-step solution

(ΔTb)PQ=Kbm\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}}=\mathrm{K}_{\mathrm{b}} \mathrm{m} 1.176=5×1M1×1000501.176=5 \times \frac{1}{\mathrm{M}_{1}} \times \frac{1000}{50} M1=85.03\mathrm{M}_{1}=85.03 (ΔTb)PQ2=5×1M2×100050=0.689\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}_{2}}=5 \times \frac{1}{\mathrm{M}_{2}} \times \frac{1000}{50}=0.689 M2=145.13\mathrm{M}_{2}=145.13 Let molar mass of P&Q\mathrm{P} \& \mathrm{Q} are MP\mathrm{M}_{\mathrm{P}} and MQ\mathrm{M}_{\mathrm{Q}} respectively MP+MQ=85.03\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}=85.03 MP+2MQ=145.13\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{Q}}=145.13 Mp=24.93≈25\mathrm{M}_{\mathrm{p}}=24.93 \approx 25 MQ=60.1≈60\mathrm{M}_{\mathrm{Q}}=60.1 \approx 60

Answer key and solution verified before publishing.

Practise Solutions and Colligative Properties

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2026
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Abnormal Colligative Properties - van't Hoff Factor
Elements P and Q form two types of non-volatile, non-ionizable… | JEE Main 2026 PYQ with Solution · DhiX AI