The chapter Solutions in Class 12 Chemistry plays a crucial role in board exams as well as competitive exams like NEET and JEE. Mastery of calculations is essential, which is why solutions numerical problems class 12 chemistry form the backbone of exam preparation. These solutions numerical problems class 12 chemistry problems test an aspirant’s understanding of concentration terms, colligative properties, and solution behavior under different conditions.

One of the first areas covered in solutions numerical problems class 12 chemistry is concentration expressions such as molarity, molality, mole fraction, and mass percentage. Numerical problems often involve converting one form of concentration into another. For example, aspirants may be asked to calculate molarity given mass, volume, and molar mass, or determine mole fraction from given masses of solute and solvent.

A major portion of solutions numerical problems class 12 chemistry focuses on colligative properties. These properties depend on the number of solute particles rather than their nature. Common numericals involve lowering of vapour pressure, elevation in boiling point, depression in freezing point, and osmotic pressure. Aspirants must clearly understand formulas like ΔTb = Kb × m and ΔTf = Kf × m to solve such questions accurately.

Another important topic in solutions numerical problems class 12 chemistry is the Van’t Hoff factor. Problems based on abnormal molar mass arise when solutes undergo association or dissociation. These numericals require careful identification of whether the solute is associating or dissociating and to what extent. Missing this step often leads to incorrect answers, even if calculations are done correctly.

Raoult’s law numericals also appear frequently in solutions numerical problems class 12 chemistry. Aspirants may be required to calculate vapour pressure of solutions, mole fraction in vapour phase, or relative lowering of vapour pressure. These questions usually involve ideal solutions and require proper application of partial pressure concepts.

Henry’s law based questions are another scoring section of solutions numerical problems class 12 chemistry. These problems relate gas solubility to pressure and involve constants like KH. Numericals may ask for mole fraction of gas dissolved in a liquid or the amount of gas absorbedI exam-oriented numerical.

Osmotic pressure calculations form a vital part of solutions numerical problems class 12 chemistry, especially for determining molar mass of macromolecules like proteins and polymers. Questions based on π = MRT are generally straightforward but require correct unit conversion, especially temperature and volume.

To score well in solutions numerical problems class 12 chemistry, aspirants must follow a systematic approach. First, write the given data clearly. Second, identify the correct formula. Third, substitute values with proper units. Finally, check whether the numerical result is logically reasonable. Skipping steps often leads to silly mistakes.

Regular practice of solutions numerical problems class 12 chemistry improves speed and accuracy. Solving previous years’ questions and NCERT-based numericals strengthens conceptual clarity. Many exam questions are directly framed from textbook examples with minor modifications.

Teachers often emphasize solutions numerical problems class 12 chemistry because they carry predictable marks. With consistent practice, aspirants can easily secure full marks from this chapter. Numerical problems are less subjective and reward clear understanding and precision.

30 MCQs on Solutions Numerical Problems Class 12 Chemistry

 1. The vapour pressure of pure benzene and toluene are 160 and 60 mm Hg respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is

Options:
A. 0.073
B. 0.027
C. 0.27
D. 0.73
Answer: D

2. Partial vapour pressure of a solution component is directly proportional to its mole fraction. This statement is known as

Options:
A. Raoult’s law
B. Distribution law
C. Henry’s law
D. Ostwald’s dilution law
Answer:  A

3.

Assertion: If a liquid solute more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase.
Reason: In the presence of a more volatile solute, only solute forms vapours.
Options:
A. Both correct and reason explains assertion
B. Both correct but reason does not explain
C. Assertion correct, reason incorrect
D. Both incorrect
Answer:  C

4. Lowering of vapour pressure is highest for

Options:
A. 0.1 M BaCl₂
B. 0.1 M glucose
C. 0.1 M MgSO₄
D. 0.1 M urea
Answer:  A

5. 12 g of a non-volatile solute dissolved in 108 g of water produces relative lowering of vapour pressure of 0.1. Molecular mass of solute is

Options:
A. 80
B. 60
C. 20
D. 40
Answer: C

6. 18 g of glucose is dissolved in 178.2 g of water. Vapour pressure of solution at 100°C is

Options:
A. 767.6 mm Hg
B. 760 mm Hg
C. 752.4 mm Hg
D. 745.4 mm Hg
Answer: C

7. Vapour pressure of benzene is 1 bar. On adding 2 g solute to 39 g benzene, vapour pressure becomes 0.8 bar. Molar mass of solute is

Options:
A. 32
B. 16
C. 64
D. 48
Answer:  B

8. 0.5 molal solute in benzene shows ΔTf = 2 K. Kf = 5. % association if solute dimerises is

Options:
A. 40%
B. 60%
C. 50%
D. 80%
Answer:  A

9. Molar enthalpy change for melting of ice is 6 kJ/mol. Internal energy change is

Options:
A. 6 − RT/1000
B. 6
C. 6 + RT/1000
D. 6RT/1000
Answer:  B

10. 18 g glucose dissolved in 90 g water. Relative lowering of vapour pressure equals

Options:
A. 0.6
B. 0.2
C. 5.1
D. 0.02
Answer: D

11. Vapour pressure of acetone drops from 185 to 183 torr on adding 1.2 g solute to 100 g acetone. Molar mass of solute is

Options:
A. 32
B. 64
C. 128
D. 488
Answer:  B

12. Vapour pressure drops from 0.80 bar to 0.60 bar on adding solute. Mole fraction of solute is

Options:
A. 0.75
B. 0.5
C. 0.25
D. 1.5
Answer:  C

13. To observe ΔTb = 0.05°C, amount of solute (M = 100) added to 100 g water (Kb = 0.5) is

Options:
A. 2 g
B. 0.5 g
C. 1 g
D. 0.75 g
Answer:  C

14. Degree of association of P is

Options:
A. 80%
B. 60%
C. 75%
D. 65%
Answer: A

15. An aqueous solution freezes at −0.125°C. Mass of water is

Options:
A. 300 g
B. 600 g
C. 500 g
D. 400 g
Answer:  D

16. Correct relation between vapour pressure lowering and elevation in boiling point is

Options:
A. (P0−Ps)/P0=ΔTb/Kb×M(P⁰ − Ps)/P⁰ = ΔTb/Kb × M(P0−Ps)/P0=ΔTb/Kb×M
B. (P0−Ps)/P0=Kb/ΔTb×M(P⁰ − Ps)/P⁰ = Kb/ΔTb × M(P0−Ps)/P0=Kb/ΔTb×M
C. (P0−Ps)/P0=Kb/ΔTb×M/1000(P⁰ − Ps)/P⁰ = Kb/ΔTb × M/1000(P0−Ps)/P0=Kb/ΔTb×M/1000
D. (P0−Ps)/P0=ΔTb/Kb×M/1000(P⁰ − Ps)/P⁰ = ΔTb/Kb × M/1000(P0−Ps)/P0=ΔTb/Kb×M/1000
Answer: D

17. Correct formulation of complex is

Options:
A. [Pt(NH₃)₄Cl₃]Cl
B. [Pt(NH₃)₄Cl₂]Cl₂
C. [Pt(NH₃)₄Cl]Cl₂
D. [Pt(NH₃)₄Cl₄]
Answer: B

18. Pressure cooker reduces cooking time because

Options:
A. Heat distributed evenly
B. Boiling point increases
C. Food is crushed
D. Chemical change occurs
Answer:  B

19. Vapour pressure of solution at 25°C is

Options:
A. 2.376
B. 24.76
C. 23.76
D. 0.097
Answer:  C

20. Vapour pressure of water containing glucose is

Options:
A. 17.439 mm Hg
B. 17.535 mm Hg
C. 17.503 mm Hg
D. 34.973 mm Hg
Answer:  A

21. Ratio of vapour pressure lowering for BaCl₂ : NaCl : Al₂(SO₄)₃ is

Options:
A. 3 : 2 : 5
B. 5 : 2 : 3
C. 5 : 3 : 2
D. 2 : 3 : 5
Answer: A

22. Depression of freezing point is

Options:
A. 7.9 K
B. 2.5 K
C. 6.6 K
D. 2.2 K
Answer:  D

23. Henry’s law constant for O₂ in water is

Options:
A. 46.82 atm
B. 43.86 atm
C. 58.44 atm
D. 74.86 atm
Answer: A

24. Partial vapour pressure of benzene is

Options:
A. 50
B. 25
C. 37.5
D. 53.5
Answer:  A

25. Ka of acetic acid is

Options:
A. 4.76 × 10⁻⁵
B. 4 × 10⁻⁵
C. 8 × 10⁻⁵
D. 2 × 10⁻⁵
Answer:  B

26. Vapour pressure at 100°C is

Options:
A. 7.6
B. 76
C. 759
D. 752.4
Answer:  D

27. Boiling point of sucrose solution is

Options:
A. 98.98°C
B. 101.02°C
C. 95.16°C
D. 100.02°C
Answer:  B

28. Mass of urea required is

Options:
A. 6.0 g
B. 3.0 g
C. 0.3 g
D. 0.6 g
Answer:  A

29. Vapour pressure of ideal solution is

Options:
A. 156
B. 145
C. 150
D. 108
Answer: A

30. Freezing point of 1% glucose solution is

Options:
A. 272.898 K
B. 0.102°C
C. 273 K
D. 0.108°C
Answer:  A

Conclusion on Solutions Numerical Problems Class 12 Chemistry

In conclusion, solutions numerical problems class 12 chemistry are essential for building confidence in physical chemistry. A strong grip on formulas, clarity in concepts, and daily practice ensure success. By mastering solutions numerical problems class 12 chemistry, aspirants not only improve exam performance but also develop analytical skills useful in higher studies.