
Important Questions Class 12 Chemistry Chapter 1 covers the chapter Solutions, one of the most important and scoring units in the Class 12 Chemistry syllabus. Since Chemistry is a high-weightage subject in board exams, students must thoroughly practice Solution Class 12 Important Questions to strengthen both conceptual clarity and numerical-solving speed.
The chapter includes key topics such as types of solutions, concentration terms, Raoult’s Law, colligative properties, abnormal molar mass, and van’t Hoff factor. These concepts are frequently tested through theory-based questions, numericals, and MCQs.
The chapter Solutions is highly numerical-based and requires strong conceptual clarity. Students must practice a variety of Solution Class 12 Chemistry Important Question types, including theoretical definitions, derivations, and calculation-based problems. Board exams often include questions on Raoult’s Law, relative lowering of vapour pressure, osmotic pressure, and depression in freezing point, making regular practice essential.
Below are exam-oriented Solution Class 12 Important Questions covering short answers, long answers, derivations, and numericals.
1. Mole fraction of glycerine C3H5(OH)3 in solution
containing 36 g of water and 46 g of glycerine is
(1) 0.46
(2) 0.40
(3) 0.20
(4) 0.36
Solution: (3) 0.20
2. (i) Define Osmotic Pressure.
Solution:
iii) If a pressure larger than the osmotic pressure is applied to the solution side, the direction of osmosis gets reversed (i.e. now the pure solvent flows out of the solution through the semi permeable membrane). This phenomenon is called reverse osmosis. It is used in desalination of sea water
3. (a) Define molal depression constant.
(b) 0.4 g of non–electrolyte dissolved in 20 g of benzene lowers its freezing point by 0.75 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute. [SAY2020]
Sol. (a)
It is the depression of freezing point for 1 molal solution.
4. The value of i (van’t Hoff factor) for aqueous solution of KCl is close to
(1) 3
(2) 1
(3) 2
(4) ½
Sol. (2)
5. (i) The vapour pressure of pure liquids A and B are 400 mm of Hg and 600 mm of Hg respectively. Calculate vapour pressure of the solution in which mole fraction of B is 0.4.
(ii) Which of the following is true for an ideal solution?
(1) ∆Hmix > 0
(2) ∆Hmix = 0
(3) ∆Vmix > 0
(4) ∆Hmix < 0
Sol.
6. Dissolution of a gas in liquid is a process
(1) with increase in enthalpy
(2) with no change in enthalpy
(3) with decrease in enthalpy
(4) for which enthalpy change cannot be predicted
Sol. (3) with decrease in enthalpy
7. a) Number of moles of the solute per kilogram of the solvent is:
(1) Mole fraction
(2) Molality
(3) Molarity
(4) Molar mass
b) 'The extent to which a solute is dissociated or associated can be expressed by Van't Hoff factor.' Substantiate the statement
c) The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A nonvolatile, non– electrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g mol–1), vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?
Sol.
7. (a) (1)
(b) van’t Hoff factor,
If the value of i < 1, association occurs and if the value of i > 1, dissociation occurs.
8. a) Henry’s law is related to solubility of a gas in liquid.
(i) State Henry’s law.
(ii) Write any two applications of Henry’s law.
b) 1000cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300K is found to be 2.57 × 10–3 bar. Calculate the molar mass of the protein. (R = 0.083 L bar/K/mol). [March 2017]
Sol.
(a) (i) Henry’s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas
Or, at constant temperature, the partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. Mathematically, p = KH.x (where p is the partial pressure, x is the mole fraction and KH is the Henry’s law constant).
(ii) In the preparation of soda water the bottle is sealed at high pressure, A medical condition known as Bends in Scuba divers A medical condition known as Anoxia in people living at high altitudes or climbers)
(b) Govem V = 1000 cm3 = 1L,
w2 = 1.26g, π = 2.57 × 10–3 bar,
T = 300 K, R = 0.083 L bar K–1 mol–1
Sol. The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a particular composition. E.g. 95% ethanol solution by volume
10. A 5% solution (by mass) of cane sugar (C12H22O11) in water has a freezing point of 271K. Calculate the freezing point of 5% (by mass) solution of glucose (C6H12O6) in water. Freezing point of pure water is 273.15 K. [March 2019]
Sol.
We know that
5% cane sugar (C12H22O11) solution by mass means 5g cane sugar is present in 100g solution. So, Mass of cane sugar (w2) = 5g Mass of solvent (w1) = 100 – 5 = 95g
Molar mass of cane sugar (M2) = 342gmol–1 freezing point of solution (Tf) = 271K freezing point of pure water (Tf°) 273.15K.
So ΔTf = Tf° – Tf = 273.15 – 271 = 2.15 K
5% glucose solution by mass means 5g glucose is present in 100g solution.
So, Mass of glucose (w2) = 5g Mass of slovent
(w1) = 100 – 5 = 95g Molar mass of glucose
(M2) = 180gmol–1 Kf 13.97 K molal–1
Therefore, freezing point of solution, Tf = Tf° – ΔTf = 273.15 – 4.085 = 269.065K
With limited time left before the Chemistry board exam, follow these quick strategies:
Revise All Formulas Daily: Concentration terms, Raoult’s Law, osmotic pressure formula, van’t Hoff factor.
Practice Numericals Repeatedly: Most marks come from calculation-based questions.
Understand Concepts, Don’t Memorize Blindly: Especially abnormal molar mass and association/dissociation cases.
Solve Previous Year Questions: Many numerical patterns repeat.
Avoid New Topics: Focus only on revision and strengthening weak areas.
Time Yourself: Practice solving numericals within limited time to simulate exam conditions.