Important Questions Class 12 Chemistry Chapter 1: Solutions

Solutions are homogeneous mixtures of two or more components with uniform composition and properties throughout the mixture.
Important Questions Class 12 Chemistry Chapter 1 help students revise concentration terms, laws, colligative properties and numericals.

Class 12 Chemistry Chapter 1 builds the base for many physical chemistry numericals in the 2026-27 board exam. The chapter explains types of solutions, concentration terms, solubility, Henry’s law, Raoult’s law, ideal and non-ideal solutions, azeotropes, colligative properties and abnormal molar mass. Students should practise mole fraction, molarity, molality, vapour pressure, boiling point elevation, freezing point depression, osmotic pressure and van’t Hoff factor carefully. This chapter is important because it appears through MCQs, assertion-reason questions, short numericals, case-based reasoning and long calculation-based questions.

Key Takeaways

• Solutions: A solution is a homogeneous mixture of two or more components.
• Henry’s Law: Gas solubility increases with pressure at constant temperature.
• Raoult’s Law: Vapour pressure of a volatile component is proportional to its mole fraction.
• Colligative Properties: These depend on the number of solute particles, not their nature.

Important Questions Class 12 Chemistry Chapter 1 Structure 2026-27

Section A: MCQs from Important Questions Class 12 Chemistry Chapter 1

Section A tests definitions, laws, formulas and quick conceptual checks. Solutions class 12 important questions in this section usually cover solution types, concentration units, Henry’s law, Raoult’s law and colligative properties.

Q1. A solution is a:

1. Heterogeneous mixture
   b. Homogeneous mixture
   c. Pure compound only
   d. Suspension only

Answer: b. Homogeneous mixture

A solution has uniform composition and properties throughout.

Q2. In a solution, the component present in the largest quantity is generally called:

1. Solute
   b. Solvent
   c. Product
   d. Catalyst

Answer: b. Solvent

The solvent usually determines the physical state of the solution.

Q3. A binary solution contains:

1. One component
   b. Two components
   c. Three components
   d. No solvent

Answer: b. Two components

A binary solution contains one solvent and one solute.

Q4. Which concentration unit depends on temperature?

1. Mass percentage
   b. Mole fraction
   c. Molality
   d. Molarity

Answer: d. Molarity

Molarity depends on volume, and volume changes with temperature.

Q5. Molality is defined as:

1. Moles of solute per litre of solution
   b. Moles of solute per kilogram of solvent
   c. Mass of solute per 100 mL solution
   d. Mole fraction of solvent

Answer: b. Moles of solute per kilogram of solvent

Molality is independent of temperature because it uses mass of solvent.

Q6. Mole fraction of all components in a solution adds up to:

1. 0
   b. 1
   c. 10
   d. 100

Answer: b. 1

The sum of mole fractions of all components is unity.

Q7. Henry’s law is written as:

1. p = KH x
   b. p = x p°
   c. ΔTb = Kb m
   d. Π = C R T

Answer: a. p = KH x

Henry’s law relates gas pressure to mole fraction of the gas in solution.

Q8. Higher value of Henry’s law constant KH means:

1. Higher gas solubility
   b. Lower gas solubility
   c. No effect on solubility
   d. Complete dissociation

Answer: b. Lower gas solubility

At a given pressure, higher KH gives lower mole fraction of dissolved gas.

Q9. Gas solubility in liquids generally:

1. Increases with temperature
   b. Decreases with temperature
   c. Is independent of pressure
   d. Becomes zero under pressure

Answer: b. Decreases with temperature

Dissolution of gas is usually exothermic, so solubility falls with temperature rise.

Q10. Raoult’s law for a volatile component is:

1. p = KH x
   b. p = x p°
   c. πV = nRT only for gases
   d. ΔTf = iKb m

Answer: b. p = x p°

Partial vapour pressure equals mole fraction multiplied by vapour pressure of pure component.

Q11. Ideal solutions obey Raoult’s law:

1. Only at low concentration
   b. Over the entire concentration range
   c. Only at high pressure
   d. Only when solute is ionic

Answer: b. Over the entire concentration range

Ideal solutions also have zero enthalpy and volume change on mixing.

Q12. For an ideal solution:

1. ΔmixH = 0 and ΔmixV = 0
   b. ΔmixH > 0 and ΔmixV > 0
   c. ΔmixH < 0 and ΔmixV < 0
   d. ΔmixH = 0 and ΔmixV > 0

Answer: a. ΔmixH = 0 and ΔmixV = 0

No heat or volume change occurs during ideal mixing.

Q13. Colligative properties depend on:

1. Nature of solute particles
   b. Number of solute particles
   c. Colour of solute
   d. Shape of container

Answer: b. Number of solute particles

Colligative properties depend on particle count in solution.

Q14. Which is a colligative property?

1. Colour
   b. Odour
   c. Osmotic pressure
   d. Viscosity only

Answer: c. Osmotic pressure

Osmotic pressure depends on the number of solute particles.

Q15. Assertion: Aquatic species are more comfortable in cold water than warm water.

Reason: Solubility of oxygen in water decreases with rise in temperature.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

Cold water contains more dissolved oxygen than warm water.

Q16. Assertion: Ethanoic acid shows abnormal molar mass in benzene.

Reason: Ethanoic acid molecules associate in benzene through hydrogen bonding.

1. Both Assertion and Reason are true, and Reason explains Assertion
   b. Both are true, but Reason does not explain Assertion
   c. Assertion is true, Reason is false
   d. Assertion is false, Reason is true

Answer: a. Both Assertion and Reason are true, and Reason explains Assertion

Association reduces the number of particles and increases observed molar mass.

Section B: Very Short Answer Questions from Class 12 Chemistry Chapter 1 Important Questions

Section B questions usually test definitions and direct formulas. Write the formula first when the question is numerical.

Q17. What is meant by concentration of solutions class 12?

Concentration of solutions class 12 means the amount of solute present in a given amount of solution or solvent.

It can be expressed as mass percentage, volume percentage, ppm, mole fraction, molarity or molality.

Q18. Define molarity and molality.

Molarity is the number of moles of solute dissolved in one litre of solution.

Molarity = moles of solute / volume of solution in litre.

Molality is the number of moles of solute dissolved in one kilogram of solvent.

Molality = moles of solute / mass of solvent in kg.

Q19. Why is molality preferred over molarity in some calculations?

Molality is preferred because it does not depend on temperature.

Molality uses mass of solvent, and mass does not change with temperature. Molarity uses volume of solution, which changes with temperature.

Q20. State Henry’s law class 12 chemistry.

Henry’s law class 12 chemistry states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution.

Formula:

p = KH x

Here, p is partial pressure, KH is Henry’s law constant and x is mole fraction of the gas.

Q21. What are isotonic, hypertonic and hypotonic solutions?

Isotonic solutions have the same osmotic pressure at a given temperature.

A hypertonic solution has higher solute concentration than the solution inside a cell, so water moves out of the cell. A hypotonic solution has lower solute concentration, so water moves into the cell.

Section C: Short Answer Questions from Solutions Class 12 Important Questions

Section C usually includes short explanations and medium numericals. Use correct units and show substitution clearly.

Q22. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

Molar mass of NaOH = 40 g mol⁻¹.

Moles of NaOH = 5 / 40 = 0.125 mol.

Volume of solution = 450 mL = 0.450 L.

Molarity = moles of solute / volume in litre.

Molarity = 0.125 / 0.450

Molarity = 0.278 M.

Q23. Calculate the molality of 2.5 g ethanoic acid in 75 g benzene.

Molar mass of ethanoic acid = 60 g mol⁻¹.

Moles of ethanoic acid = 2.5 / 60 = 0.0417 mol.

Mass of benzene = 75 g = 0.075 kg.

Molality = moles of solute / mass of solvent in kg.

Molality = 0.0417 / 0.075

Molality = 0.556 mol kg⁻¹.

Q24. Explain solubility of a gas in a liquid using Henry’s law.

Henry’s law says that gas solubility increases with pressure at constant temperature.

When pressure above the liquid increases, more gas particles strike the surface of the solution. More gas particles enter the liquid phase until a new equilibrium is reached.

This is why carbon dioxide is sealed under high pressure in soda bottles. It increases the amount of dissolved CO₂.

Q25. Give three applications of Henry’s law.

Three applications of Henry’s law are:

1. CO₂ is dissolved in soft drinks under high pressure.
2. Scuba divers use air diluted with helium to avoid bends.
3. At high altitudes, low oxygen pressure causes anoxia.

Henry’s law explains gas solubility changes in each case.

Q26. Distinguish between ideal and non-ideal solutions class 12.

Non-ideal solutions show positive or negative deviation from Raoult’s law.

Q27. Explain positive and negative deviation from Raoult’s law.

Positive deviation occurs when A-B interactions are weaker than A-A and B-B interactions.

Molecules escape more easily, so vapour pressure becomes higher than predicted. Ethanol and acetone show positive deviation.

Negative deviation occurs when A-B interactions are stronger than A-A and B-B interactions.

Molecules escape less easily, so vapour pressure becomes lower than predicted. Chloroform and acetone show negative deviation.

Q28. Calculate the boiling point of water when 18 g glucose is dissolved in 1 kg water.

Molar mass of glucose = 180 g mol⁻¹.

Moles of glucose = 18 / 180 = 0.1 mol.

Mass of water = 1 kg.

Molality = 0.1 mol kg⁻¹.

For water, Kb = 0.52 K kg mol⁻¹.

Elevation of boiling point = Kb × m.

ΔTb = 0.52 × 0.1 = 0.052 K.

Boiling point of solution = 373.15 + 0.052 = 373.202 K.

Section D: Case-Based Questions from Class 12 Chemistry Solutions Important Questions

Case-based questions from Solutions usually combine a situation with a law or colligative property. Identify the principle first, then apply the formula.

Q29. Case Study: Scuba diving and dissolved gases

A scuba diver breathes air under high pressure underwater. At high pressure, more nitrogen dissolves in blood. If the diver rises quickly, pressure decreases and nitrogen bubbles may form in blood, causing bends.

Q29(a). Which law explains this situation?

Henry’s law explains this situation.

Gas solubility in liquid increases with pressure.

Q29(b). Why does nitrogen dissolve more underwater?

Underwater pressure is higher.

Higher pressure increases the solubility of nitrogen in blood.

Q29(c). Why are bends dangerous?

Bends are dangerous because nitrogen bubbles can block capillaries.

This can cause pain and may become life-threatening.

Q29(d). How is this problem reduced?

Scuba tanks are filled with air diluted with helium.

Helium reduces the harmful effect of high nitrogen concentration.

Q30. Case Study: Intravenous saline and osmosis

A patient is given normal saline through an intravenous injection. Normal saline contains 0.9% mass/volume sodium chloride and has osmotic pressure similar to blood cell fluid.

Q30(a). What type of solution is normal saline for blood cells?

Normal saline is isotonic for blood cells.

It has osmotic pressure similar to the fluid inside blood cells.

Q30(b). What happens if blood cells are placed in hypertonic solution?

Water moves out of the cells.

The cells shrink due to osmosis.

Q30(c). What happens if blood cells are placed in hypotonic solution?

Water moves into the cells.

The cells swell because the outside solution has lower salt concentration.

Q30(d). Which colligative property is involved?

Osmotic pressure is involved.

It depends on the number of solute particles in solution.

Section E: Long Answer Questions from Important Questions Class 12 Chemistry Chapter 1

Long answers from this chapter usually ask for full explanations or multi-step numericals. Write definitions, formulas, substitution and final units clearly.

Q31. Explain colligative properties class 12 with examples.

Colligative properties class 12 are properties that depend on the number of solute particles and not on their nature.

The four colligative properties are:

1. Relative lowering of vapour pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure

When a non-volatile solute is added to a volatile solvent, vapour pressure decreases. This causes boiling point elevation and freezing point depression.

Osmotic pressure arises when solvent flows through a semipermeable membrane from dilute solution to concentrated solution.

Examples include salt lowering the freezing point of ice, sugar preserving fruits and saline being isotonic with blood.

Q32. Explain osmotic pressure class 12 and reverse osmosis.

Osmosis is the flow of solvent through a semipermeable membrane from lower solute concentration to higher solute concentration.

Osmotic pressure is the excess pressure needed to stop osmosis.

Formula:

Π = C R T

or

ΠV = nRT

Here, Π is osmotic pressure, C is molarity, R is gas constant and T is temperature.

Osmotic pressure is useful for finding molar masses of proteins and polymers because it can be measured at room temperature.

Reverse osmosis occurs when pressure greater than osmotic pressure is applied on the solution side.

In reverse osmosis, pure solvent moves out of the solution. This method is used for desalination of sea water.

Q33. Explain abnormal molar mass and van’t Hoff factor class 12.

Abnormal molar mass occurs when the experimentally determined molar mass is different from the normal expected molar mass.

This happens due to association or dissociation of solute particles.

In association, particles combine, so the number of particles decreases. The observed molar mass becomes higher than the normal value. Ethanoic acid dimerises in benzene due to hydrogen bonding.

In dissociation, particles split into ions, so the number of particles increases. The observed molar mass becomes lower than the normal value. KCl dissociates into K⁺ and Cl⁻ in water.

Van’t Hoff factor i accounts for association or dissociation.

Formula:

i = normal molar mass / abnormal molar mass

It can also be written as:

i = observed colligative property / calculated colligative property

Modified formulas are:

Elevation of boiling point = i Kb m

Depression of freezing point = i Kf m

Osmotic pressure = i nRT / V

For association, i is less than 1. For dissociation, i is greater than 1.

Formula-Based Revision for Important Questions Class 12 Chemistry Chapter 1

Important questions class 12 chemistry chapter 1 should be revised through formulas and applications. Keep units clear while solving numericals.

Concentration of Solutions Class 12

Mass percentage = Mass of component × 100 / Total mass of solution.

Volume percentage = Volume of component × 100 / Total volume of solution.

Parts per million = Number of parts of component × 10⁶ / Total number of parts.

Mole fraction = Moles of component / Total moles of all components.

Molarity = Moles of solute / Volume of solution in litre.

Molality = Moles of solute / Mass of solvent in kg.

Henry’s Law Class 12 Chemistry

p = KH x

Here:

p = partial pressure of gas

KH = Henry’s law constant

x = mole fraction of gas in solution

Higher KH means lower gas solubility.

Raoult’s Law Class 12 Chemistry

For a volatile component:

p = x p°

Here:

p = partial vapour pressure

x = mole fraction

p° = vapour pressure of pure component

For total vapour pressure of a binary solution:

ptotal = p1 + p2

Relative Lowering of Vapour Pressure

Relative lowering of vapour pressure = (p° - p) / p°

For a dilute solution:

(p° - p) / p° = n2 / n1

Here, n2 is moles of solute and n1 is moles of solvent.

Elevation of Boiling Point Class 12

ΔTb = Kb m

For abnormal molar mass:

ΔTb = i Kb m

Here, Kb is molal elevation constant and m is molality.

Depression of Freezing Point Class 12

ΔTf = Kf m

For abnormal molar mass:

ΔTf = i Kf m

Here, Kf is molal depression constant and m is molality.

Osmotic Pressure Class 12

Π = C R T

ΠV = nRT

For abnormal molar mass:

Π = i nRT / V

Van’t Hoff Factor Class 12

i = normal molar mass / abnormal molar mass.

i = observed colligative property / calculated colligative property.

For association, i < 1.

For dissociation, i > 1.

Chapter-Wise Revision for Solutions Class 12 Important Questions

Important questions class 12 chemistry chapter 1 should be revised in seven parts: solution types, concentration, solubility, laws, ideal solutions, colligative properties and abnormal molar mass.

Start with types of solutions and concentration terms. Practise mass percentage, mole fraction, molarity and molality.

Next, revise solubility. Remember that “like dissolves like” and pressure strongly affects gas solubility.

Then revise Henry’s law class 12 chemistry. Focus on soda bottles, scuba diving, cold water oxygen and high-altitude anoxia.

After that, revise Raoult’s law class 12 chemistry. Understand vapour pressure of volatile liquids and non-volatile solutes.

Then revise ideal and non-ideal solutions class 12. Learn positive deviation, negative deviation and azeotropes with examples.

Next, practise colligative properties class 12. Focus on boiling point elevation, freezing point depression and osmotic pressure numericals.

Finally, revise van’t Hoff factor class 12. Association gives i less than 1, while dissociation gives i greater than 1.

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