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Solutions JEE Main MCQ & Practice Questions

Practice Solutions MCQs for JEE Main Chemistry with answers, explanations, chapter revision and related ChemNexa mock tests.

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Solutions Revision Guide for JEE Main Chemistry

Solutions deals with concentration terms, vapour pressure and colligative properties. Many exam questions are short numericals where correct units and interpretation are more important than lengthy calculation.

For JEE Main Chemistry, revise mole fraction, molarity, molality, Raoult law, ideal and non-ideal behaviour, elevation of boiling point, depression of freezing point and osmotic pressure.

Important Topics

  • Concentration terms
  • Raoult law
  • Ideal and non-ideal solutions
  • Colligative properties
  • van’t Hoff factor
  • Osmotic pressure

Important Formulae & Relationships

  • Molarity M = moles of solute / volume of solution (L)
  • Molality m = moles of solute / mass of solvent (kg)
  • ΔTb = iKb m
  • ΔTf = iKf m
  • π = iCRT

Use formulas only after checking the required units, sign convention and assumptions for the question.

What to practise in Solutions

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Top 20 Solutions MCQs with Answers

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JEE Main Chemistry · Solutions · Very Very Hard

1. Raoult's law for a volatile component i in an ideal solution is:

  • A. pi=pi°/xi
  • B. pi=xi/P°
  • C. pi=xiRT
  • D. pi=xi pi°
Answer: D
Explanation: Partial vapour pressure equals liquid mole fraction times pure-component vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

2. Consider: (I) molality is temperature independent; (II) ideal solutions obey Raoult's law; (III) ΔTf=iKfm; (IV) reverse osmosis requires pressure greater than π. Which are correct?

  • A. I, III and IV only
  • B. I, II, III and IV
  • C. I and II only
  • D. II and III only
Answer: B
Explanation: All four statements are correct.

JEE Main Chemistry · Solutions · Very Very Hard

3. JEE Main advanced: Raoult's law for a volatile component i in an ideal solution is:

  • A. pi=xiRT
  • B. pi=xi pi°
  • C. pi=pi°/xi
  • D. pi=xi/P°
Answer: B
Explanation: Partial vapour pressure equals liquid mole fraction times pure-component vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

4. JEE Main advanced: Consider: (I) molality is temperature independent; (II) ideal solutions obey Raoult's law; (III) ΔTf=iKfm; (IV) reverse osmosis requires pressure greater than π. Which are correct?

  • A. I and II only
  • B. II and III only
  • C. I, III and IV only
  • D. I, II, III and IV
Answer: D
Explanation: All four statements are correct.

JEE Main Chemistry · Solutions · Very Very Hard

5. Solution-property reasoning: Raoult's law for a volatile component i in an ideal solution is:

  • A. pi=pi°/xi
  • B. pi=xi/P°
  • C. pi=xiRT
  • D. pi=xi pi°
Answer: D
Explanation: Partial vapour pressure equals liquid mole fraction times pure-component vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

6. If xsolute=0.10 in an ideal solution with nonvolatile solute, relative lowering of vapour pressure is:

  • A. 10
  • B. 0.10
  • C. 0.90
  • D. 1.10
Answer: B
Explanation: Relative lowering equals solute mole fraction.

JEE Main Chemistry · Solutions · Very Very Hard

7. JEE Main advanced: If xsolute=0.10 in an ideal solution with nonvolatile solute, relative lowering of vapour pressure is:

  • A. 0.90
  • B. 1.10
  • C. 10
  • D. 0.10
Answer: D
Explanation: Relative lowering equals solute mole fraction.

JEE Main Chemistry · Solutions · Very Very Hard

8. Molality is preferred over molarity in temperature-dependent studies because molality:

  • A. Uses solution volume
  • B. Depends strongly on expansion
  • C. Is based on masses
  • D. Is always numerically smaller
Answer: C
Explanation: Masses are essentially temperature independent.

JEE Main Chemistry · Solutions · Very Very Hard

9. Negative deviation from Raoult's law generally occurs when A-B attractions are:

  • A. Weaker than all like interactions
  • B. Absent
  • C. Purely repulsive
  • D. Stronger than A-A and B-B attractions
Answer: D
Explanation: Stronger unlike attractions lower escaping tendency and vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

10. A minimum-boiling azeotrope is associated generally with:

  • A. Perfect ideality only
  • B. Zero vapour pressure
  • C. Positive deviation from Raoult's law
  • D. Negative deviation
Answer: C
Explanation: Large positive deviations can generate maximum vapour pressure and minimum boiling point.

JEE Main Chemistry · Solutions · Very Very Hard

11. A maximum-boiling azeotrope is associated generally with:

  • A. Ideal gas behaviour
  • B. Negative deviation from Raoult's law
  • C. Positive deviation
  • D. No intermolecular attraction
Answer: B
Explanation: Large negative deviations can generate minimum vapour pressure and maximum boiling point.

JEE Main Chemistry · Solutions · Very Very Hard

12. Relative lowering of vapour pressure for a dilute nonvolatile solute equals approximately:

  • A. Mole fraction of solute
  • B. Mole fraction of solvent
  • C. Molality directly
  • D. Molar mass of solvent
Answer: A
Explanation: Raoult's law gives (p°-p)/p°=xsolute.

JEE Main Chemistry · Solutions · Very Very Hard

13. Addition of a nonvolatile solute to a solvent causes vapour pressure to:

  • A. Increase
  • B. Remain unchanged
  • C. Become infinite
  • D. Decrease
Answer: D
Explanation: The solvent mole fraction decreases, lowering its vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

14. Boiling point elevation is represented by:

  • A. ΔTb=m/Kb
  • B. ΔTb=Kb/m
  • C. ΔTb=Kb m
  • D. ΔTb=Kf m
Answer: C
Explanation: For dilute solutions, elevation is proportional to molality.

JEE Main Chemistry · Solutions · Very Very Hard

15. Freezing point depression is represented by:

  • A. ΔTf=Kf/m
  • B. ΔTf=Kf m
  • C. ΔTf=Kb m
  • D. ΔTf=m/Kf
Answer: B
Explanation: For dilute solutions, depression is proportional to molality.

JEE Main Chemistry · Solutions · Very Very Hard

16. Adding a nonvolatile solute generally causes boiling point to:

  • A. Increase
  • B. Decrease
  • C. Remain unchanged
  • D. Become 0°C
Answer: A
Explanation: Lower vapour pressure requires a higher temperature to reach external pressure.

JEE Main Chemistry · Solutions · Very Very Hard

17. For an electrolyte, osmotic pressure is:

  • A. π=Kf m
  • B. π=iCRT
  • C. π=CRT/i
  • D. π=iKb m
Answer: B
Explanation: van't Hoff factor corrects osmotic pressure.

JEE Main Chemistry · Solutions · Very Very Hard

18. JEE Main advanced: Molality is preferred over molarity in temperature-dependent studies because molality:

  • A. Is based on masses
  • B. Is always numerically smaller
  • C. Uses solution volume
  • D. Depends strongly on expansion
Answer: A
Explanation: Masses are essentially temperature independent.

JEE Main Chemistry · Solutions · Very Very Hard

19. JEE Main advanced: Negative deviation from Raoult's law generally occurs when A-B attractions are:

  • A. Purely repulsive
  • B. Stronger than A-A and B-B attractions
  • C. Weaker than all like interactions
  • D. Absent
Answer: B
Explanation: Stronger unlike attractions lower escaping tendency and vapour pressure.

JEE Main Chemistry · Solutions · Very Very Hard

20. JEE Main advanced: A minimum-boiling azeotrope is associated generally with:

  • A. Positive deviation from Raoult's law
  • B. Negative deviation
  • C. Perfect ideality only
  • D. Zero vapour pressure
Answer: A
Explanation: Large positive deviations can generate maximum vapour pressure and minimum boiling point.

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Solutions Practice Tests

Solutions Frequently Asked Questions

Which concentration unit is temperature independent?

Molality is based on mass of solvent and therefore does not change with temperature, unlike molarity which depends on solution volume.

Why is the van’t Hoff factor used?

It accounts for association or dissociation of solute particles so that observed colligative properties can be related to the effective number of dissolved particles.

How can I improve accuracy in Solutions numericals?

Convert masses, volumes and molecular masses carefully, write the chosen concentration unit explicitly and check whether association or dissociation affects the particle count.

Continue with another chapter to build connected concepts and strengthen your overall Chemistry preparation.

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