Unit 1: Solutions – Previous Year Board Questions
Note for Students: This resource provides very important questions from Unit 1: Solutions along with authentic previous years' board questions (PYQs) from CBSE, ISC, NIOS and other Board exams. Use this to master important concepts and practice questions asked in real board exams.
1 Mark Questions
CBSE 2023: Define molality and write its SI unit. (1 mark)
Molality is the number of moles of solute per kg of solvent; SI unit: mol kg−1.
CBSE 2022: State Raoult's law in brief. (1 mark)
The partial vapour pressure of each component is directly proportional to its mole fraction in an ideal solution.
CBSE 2021: What is an azeotrope? (1 mark)
A mixture with constant boiling point and composition; cannot be separated by simple distillation.
CBSE 2021: What is the Van't Hoff factor? (1 mark)
It is the ratio of actual number of solute particles present after dissociation/association to the number of formula units initially dissolved.
CBSE 2020: Name one colligative property. (1 mark)
Depression in freezing point.
2 Mark Questions
ISC 2024: A 5% aqueous solution of glucose (molar mass = 180 g/mol) is isotonic with 1.66% aqueous solution of urea. Calculate the molar mass of urea. (2 marks)
Molar mass of urea = approximately 60 g/mol (using isotonic solution relation π1 = π2)
ISC 2024: Calculate the molar mass of protein if the osmotic pressure of a 200 cm³ solution containing 1.26 g protein at 300 K is 2.57 × 10⁻³ atm. (2 marks)
Molar mass of protein ≈ 60,380 g/mol (using π = CRT formula)
ISC 2024: Calculate the maximum electrical work obtainable from Zn|Zn²⁺ || Ag⁺|Ag galvanic cell at standard conditions 298 K. Given E°(Zn²⁺/Zn) = –0.76 V and E°(Ag⁺/Ag) = +0.80 V. (2 marks)
Maximum electrical work = –301.08 kJ (using W = -nFE°cell)
CBSE 2023: Differentiate between molarity and molality. (2 marks)
Molarity is moles of solute per litre solution, depends on temperature; molality is moles of solute per kg solvent, independent of temperature.
ISC 2023: Calculate the mass of ascorbic acid (molecular mass = 176 g/mol) that should be dissolved in 155g of acetic acid to cause a depression of freezing point by 1.15K. (2 marks)
Mass of ascorbic acid needed = 8.04 g (using formula ΔTf = Kf × molality)
ISC 2023: The osmotic pressure of blood at 37°C is 8.21 atm. How much glucose in grams should be used per litre of aqueous solution for an intravenous injection so that it is isotonic with blood? (2 marks)
Mass of glucose required = 58.06 g (Using π = CRT and M = 180 g/mol)
ISC 2023: Calculate the molar mass of protein if the osmotic pressure of a 200 cm³ solution containing 1.26 g protein at 300 K is 2.57 × 10−3 atm. (2 marks)
Molar mass of protein is approximately 60,380 g/mol (using osmotic pressure formula π = CRT)
CBSE 2022: Why is osmotic pressure a preferred method for determining molar mass of macromolecules? (2 marks)
Because it can be measured at room temperature (prevents denaturation) and is precise even for very dilute solutions.
ISC 2022: Calculate the mass of ascorbic acid (molecular mass = 176 g/mol) that should be dissolved in 155 g of acetic acid to cause a depression of freezing point by 1.15 K. (2 marks)
Mass of ascorbic acid needed = 8.04 g (using ΔTf = Kf × molality; Kf for acetic acid = 3.9 K kg/mol)
ISC 2022: The osmotic pressure of blood at 37°C is 8.21 atm. How much glucose should be used per litre of aqueous solution for intravenous injection so that it is isotonic with blood? (2 marks)
Mass of glucose required = 58.06 g (using π = CRT with molar mass 180 g/mol)
ISC 2022: Calculate the molar mass of protein if the osmotic pressure of a 200 cm³ solution containing 1.26 g protein at 300 K is 2.57 × 10⁻³ atm. (2 marks)
Molar mass of protein ≈ 60,380 g/mol (from osmotic pressure formula π = CRT)
CBSE 2021: State Henry's law and give one application. (2 marks)
Henry's law: Solubility of a gas in a liquid is proportional to its partial pressure. Used in bottling soft drinks under high CO2 pressure.
CBSE 2020: Between 0.1 M NaCl and 0.1 M glucose, which will have a lower freezing point? Explain. (2 marks)
0.1 M NaCl.
NaCl dissociates to give more particles, causing more freezing point depression.
NaCl dissociates to give more particles, causing more freezing point depression.
CBSE 2020: Give an example each of minimum and maximum boiling azeotropes. (2 marks)
Minimum Boiling Azeotropes: Ethanol-Water
Maximum Boiling Azeotropes: Nitric Acid–Water.
Maximum Boiling Azeotropes: Nitric Acid–Water.
3 Mark Questions
ISC 2024: An aqueous solution containing 12.50 g of barium chloride in 1000 g of water boils at 373.0834 K. Calculate the degree of dissociation of barium chloride. (3 marks)
Degree of dissociation of BaCl₂ ≈ 85.9%
CBSE 2023: Explain the effect of temperature on solubility of gases in liquids, with an example. (3 marks)
Solubility of gases decreases with increase in temperature.
Example: CO2 is more soluble in cold water; aquatic animals survive better in cold water.
Example: CO2 is more soluble in cold water; aquatic animals survive better in cold water.
ISC 2023: When 2g of benzoic acid is dissolved in 25g of benzene, it shows a depression in freezing point equal to 1.62K. Calculate the percentage association of benzoic acid. (3 marks)
Degree of association of benzoic acid is 99.2% (Benzoic acid forms dimers in benzene)
ISC 2023: The osmotic pressure of blood at 37°C is 8.21 atm. How much glucose in grams should be used per litre of aqueous solution for an intravenous injection so that it is isotonic with blood? (Molecular wt of glucose = 180 g/mol). (3 marks)
T = 273 + 37 = 310 K
Since, π = CRT
where C is molar concentration.
8.21 = w/M x 0.082 x 310
where M = 180 g/mol
w = (8.21 x 180)/(0.082 x 310) = 58.13 gm
∴ To prepare a solution isotonic with blood, 58.13 g of glucose per litre of solution is required.
Since, π = CRT
where C is molar concentration.
8.21 = w/M x 0.082 x 310
where M = 180 g/mol
w = (8.21 x 180)/(0.082 x 310) = 58.13 gm
∴ To prepare a solution isotonic with blood, 58.13 g of glucose per litre of solution is required.
CBSE 2022: Explain positive and negative deviations from Raoult's law with suitable examples. (3 marks)
Positive Deviation: Weaker A-B interactions, higher vapour pressure than expected (e.g., ethanol–acetone);
Negative Deviation: Stronger A-B interactions, lower vapour pressure (e.g., chloroform–acetone).
Negative Deviation: Stronger A-B interactions, lower vapour pressure (e.g., chloroform–acetone).
CBSE 2022: Calculate the molality of a solution prepared by dissolving 5.8 g NaCl in 500 g water. (3 marks)
Moles NaCl = 5.8 / 58.5 = 0.1 mol;
Mass of water = 0.5 kg;
Molality = 0.1 / 0.5 = 0.2 mol kg−1.
Mass of water = 0.5 kg;
Molality = 0.1 / 0.5 = 0.2 mol kg−1.
ISC 2022: When 2 g of benzoic acid is dissolved in 25 g of benzene, it shows a depression in freezing point equal to 1.62 K. Calculate the percentage association of benzoic acid. (3 marks)
Degree of association of benzoic acid = 99.2% (Benzoic acid forms dimers in benzene)
CBSE 2021: What is abnormal molar mass? Why is it observed? (3 marks)
Molar mass that differs from expected due to association (e.g., dimerisation) or dissociation of solute in solvent.
CBSE 2020: Derive the mathematical relation for elevation of boiling point in terms of molality. (3 marks)
ΔTb = Kb × m,
where ΔTb is elevation in boiling point, Kb is molal elevation constant, m is molality.
where ΔTb is elevation in boiling point, Kb is molal elevation constant, m is molality.
5 Mark Questions
CBSE 2023: a) State and explain Raoult's law for volatile liquid solutions. b) Discuss ideal and non-ideal solutions with examples. (5 marks)
a) Raoult's law states that for a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction in that solution.
b) Ideal Solutions: An ideal solution is one that perfectly obeys Raoult's law at all concentrations and temperatures. This behavior occurs when the intermolecular forces of attraction between the molecules of different components (A-B interactions) are identical to the forces between the molecules of the pure components (A-A and B-B interactions).
ΔHmix = 0
ΔVmix = 0
Example: Benzene and Toluene
Non-Ideal Solutions: A non-ideal solution is one that does not obey Raoult's law. In these solutions, the interactions between the different molecules (A-B) are either stronger or weaker than the interactions in the pure components (A-A, B-B). This deviation leads to two different behaviors:
1. Positive Deviation from Raoult's Law: This occurs when the forces between unlike molecules (A-B) are weaker than those between like molecules (A-A, B-B). The molecules find it easier to escape the solution, leading to a vapor pressure that is higher than predicted by Raoult's law.
ΔHmix > 0
ΔVmix > 0
Example: Ethanol and Acetone.
2. Negative Deviation from Raoult's Law: This occurs when the forces between unlike molecules (A-B) are stronger than those between like molecules (A-A, B-B). The molecules are held more tightly in the solution, making them less likely to escape into the vapor phase. This results in a vapor pressure that is lower than predicted by Raoult's law.
ΔHmix < 0
ΔVmix < 0
Example: Chloroform and Acetone.
b) Ideal Solutions: An ideal solution is one that perfectly obeys Raoult's law at all concentrations and temperatures. This behavior occurs when the intermolecular forces of attraction between the molecules of different components (A-B interactions) are identical to the forces between the molecules of the pure components (A-A and B-B interactions).
ΔHmix = 0
ΔVmix = 0
Example: Benzene and Toluene
Non-Ideal Solutions: A non-ideal solution is one that does not obey Raoult's law. In these solutions, the interactions between the different molecules (A-B) are either stronger or weaker than the interactions in the pure components (A-A, B-B). This deviation leads to two different behaviors:
1. Positive Deviation from Raoult's Law: This occurs when the forces between unlike molecules (A-B) are weaker than those between like molecules (A-A, B-B). The molecules find it easier to escape the solution, leading to a vapor pressure that is higher than predicted by Raoult's law.
ΔHmix > 0
ΔVmix > 0
Example: Ethanol and Acetone.
2. Negative Deviation from Raoult's Law: This occurs when the forces between unlike molecules (A-B) are stronger than those between like molecules (A-A, B-B). The molecules are held more tightly in the solution, making them less likely to escape into the vapor phase. This results in a vapor pressure that is lower than predicted by Raoult's law.
ΔHmix < 0
ΔVmix < 0
Example: Chloroform and Acetone.
CBSE 2022: a) How is molecular mass determined using colligative properties? b) What modifications are required for electrolytes? (5 marks)
a) By measuring a colligative property (e.g., depression of freezing point) and applying the relevant formula.
b) Account for the Van’t Hoff factor (i) due to association/dissociation when calculating.
b) Account for the Van’t Hoff factor (i) due to association/dissociation when calculating.
CBSE 2021: a) What are colligative properties? Name any four. b) Why are these properties called colligative? (5 marks)
a) Colligative properties: Properties that depend only on the number of solute particles, not their nature.
Examples: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure.
b) They depend on the 'collection' (number) of particles.
Examples: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, osmotic pressure.
b) They depend on the 'collection' (number) of particles.
CBSE 2020: A 5% (by mass) solution of a non-volatile solute in benzene has a freezing point of 274.0 K. Pure benzene freezes at 278.4 K. Kf for benzene is 5.12 K kg mol−1. Calculate the molar mass of the solute. (5 marks)
ΔTf = 4.4 K; m = ΔTf/Kf = 4.4/5.12 = 0.86 mol/kg;
5 g solute in 95 g benzene: (5/0.086)x = 58.14 g/mol.
Final answer: The molar mass of the solute is approximately 58 g/mol.
5 g solute in 95 g benzene: (5/0.086)x = 58.14 g/mol.
Final answer: The molar mass of the solute is approximately 58 g/mol.
CBSE 2019: a) Define Henry's law. b) Explain with reason why deep-sea divers use a mixture of He and O2 for breathing. (5 marks)
a) Henry's law: Solubility of a gas in liquid is proportional to partial pressure above the solution.
b) At depth, nitrogen's increased solubility is narcotic; helium’s low solubility prevents this, making the mixture safe for divers.
b) At depth, nitrogen's increased solubility is narcotic; helium’s low solubility prevents this, making the mixture safe for divers.
Important Questions with Exam Name & Year
[CBSE Class 12 Chemistry Exam 2025] What are colligative properties? List the four colligative properties important in this chapter.
Colligative properties depend on the number of solute particles, not their identity.
The four are:
1) Relative lowering of vapour pressure,
2) Elevation of boiling point,
3) Depression of freezing point,
4) Osmotic pressure.
The four are:
1) Relative lowering of vapour pressure,
2) Elevation of boiling point,
3) Depression of freezing point,
4) Osmotic pressure.
[CBSE Board 2024] Explain Raoult's law and its significance in solution behaviour.
Raoult’s law states: Partial vapour pressure of each volatile component in a solution is proportional to its mole fraction.
It helps predict vapour pressures and deviations (positive/negative), explaining ideal and non-ideal solutions.
It helps predict vapour pressures and deviations (positive/negative), explaining ideal and non-ideal solutions.
[CBSE Class 12 Annual Exam 2023] Describe the Van't Hoff factor and its importance in colligative property calculations.
The Van’t Hoff factor (i) corrects colligative property calculations by considering solute dissociation or association.
i = 1 for non-electrolytes
i > 1 for electrolytes (dissociation)
i < 1 for association (e.g., acetic acid).
i = 1 for non-electrolytes
i > 1 for electrolytes (dissociation)
i < 1 for association (e.g., acetic acid).
[CBSE Midterm 2025] How does Henry's law explain the solubility of gases in liquids? Give an application.
Henry's law states solubility of a gas is proportional to its partial pressure above the liquid.
Application: Carbonated drinks are bottled under high CO2 pressure to increase solubility.
Application: Carbonated drinks are bottled under high CO2 pressure to increase solubility.
[CBSE Compartment Exam 2024] What is an azeotrope? Give examples of minimum and maximum boiling azeotropes.
An azeotrope is a mixture boiling at constant temperature and composition.
Minimum boiling: ethanol-water (95.4%),
Maximum boiling: nitric acid-water.
Minimum boiling: ethanol-water (95.4%),
Maximum boiling: nitric acid-water.
[CBSE Class 12 Chemistry Board 2025] Why is osmotic pressure considered the best method for determining molecular masses of macromolecules?
Because it can measure molecular masses in very dilute solutions without heating, preventing molecule damage.
[CBSE Board 2023] Explain the effect of temperature on the solubility of gases in liquids.
Solubility decreases with increasing temperature as gas molecules gain energy and escape from liquid (e.g., aquatic animals prefer cold water).
[CBSE Class 12 Chemistry Exam 2024] Why are colligative properties important in real-world applications? Give one example.
They help in antifreeze formulation (freezing point depression) and in medical treatments like dialysis (osmotic pressure).
[CBSE Board 2023] Calculate molarity, molality, and mole fraction for a solution containing 46 g ethanol in 180 g water.
Basic steps:
- Moles ethanol = mass/molar mass = 0.9985 mol,
- Moles of Water = mass/molar mass = 9.9889 mol,
- Molality = moles solute/kg solvent = 5.5471 m,
- Mole Fraction of Ethanol = 0.0909,
- Mole Fraction of Water = 0.9091,
- Total moles for mole fraction = 10.9874 mol,
- Density of Ethanol: 0.789 g/mL,
- Volume of Ethanol = 58.30 ml,
- Volume of Water = 180 ml,
- Total Volume of Solution = 238.30 ml = 0.2383 ltr,
- Molarity = moles of ethanol/volume in solution in ltr. = 41900 M.
- Moles ethanol = mass/molar mass = 0.9985 mol,
- Moles of Water = mass/molar mass = 9.9889 mol,
- Molality = moles solute/kg solvent = 5.5471 m,
- Mole Fraction of Ethanol = 0.0909,
- Mole Fraction of Water = 0.9091,
- Total moles for mole fraction = 10.9874 mol,
- Density of Ethanol: 0.789 g/mL,
- Volume of Ethanol = 58.30 ml,
- Volume of Water = 180 ml,
- Total Volume of Solution = 238.30 ml = 0.2383 ltr,
- Molarity = moles of ethanol/volume in solution in ltr. = 41900 M.
[CBSE Annual Exam 2025] What mistakes are commonly made by students in numerical problems on colligative properties?
Ignoring the Van't Hoff factor (i) for electrolytes, leading to incorrect molecular mass or property calculations.