Solutions play a central role in chemistry and in our daily life. From the air we breathe (a gaseous solution of oxygen in nitrogen) to the salt water in oceans, solutions surround us everywhere. In Class 12 Chemistry, the chapter “Solutions” explains different types of solutions, their concentration methods, solubility, colligative properties, and the mathematical relationships used to describe them.
This chapter is not only important for board examinations but also forms the foundation for competitive exams like JEE and NEET. Let us study each concept in detail.
Class 12 Chemistry Solutions Notes | Definition, Types, Laws, Colligative Properties & MCQs
Definition of Solutions
A solution is a homogeneous mixture of two or more substances whose composition can vary within certain limits.
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Solvent: The component present in a larger amount and dissolves the other component(s).
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Solute: The component present in a smaller amount and gets dissolved in the solvent.
Example: In salt solution, water is the solvent and salt is the solute.
Types of Solutions
Solutions can be classified based on the physical state of solute and solvent.
On the Basis of Physical State
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Gaseous Solutions
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Gas in Gas → Air (O₂ + N₂)
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Gas in Liquid → CO₂ in soda water
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Gas in Solid → Hydrogen in palladium
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Liquid Solutions
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Liquid in Liquid → Alcohol in water
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Solid in Liquid → Sugar in water
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Gas in Liquid → Oxygen dissolved in water
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Solid Solutions
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Solid in Solid → Alloys (brass: Zn + Cu)
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Gas in Solid → Hydrogen in metals
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Liquid in Solid → Amalgams
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Concentration of Solutions
The amount of solute present in a given quantity of solution is called its concentration. Several units are used to express concentration:
Mass Percentage (% by mass)
Mass %=Mass of soluteMass of solution×100\text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100
Volume Percentage (% by volume)
Volume %=Volume of soluteVolume of solution×100\text{Volume \%} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100
Mass by Volume Percentage
Mass of solute (g)Volume of solution (mL)×100\frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100
Parts per Million (ppm)
Used for very low concentrations.
ppm=Mass of soluteMass of solution×106\text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6
Mole Fraction (χ)
χA=nAnA+nB\chi_A = \frac{n_A}{n_A + n_B}
where nA,nBn_A, n_B are moles of solute and solvent.
Molarity (M)
M=Moles of soluteVolume of solution in litreM = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}}
Molality (m)
m=Moles of soluteMass of solvent in kgm = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}
Solubility
Solubility is the maximum amount of solute that can be dissolved in 100 g of solvent at a given temperature and pressure.
Factors Affecting Solubility
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Nature of solute and solvent – Like dissolves like.
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Temperature – Solubility of solids in liquids generally increases with temperature.
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Pressure – Solubility of gases in liquids increases with pressure (Henry’s Law).
Henry’s Law
The solubility of a gas in a liquid is directly proportional to the pressure of the gas over the liquid.
p=kH⋅xp = k_H \cdot x
where:
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pp = partial pressure of gas
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xx = mole fraction of gas in solution
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kHk_H = Henry’s law constant
Applications:
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Bottling of soft drinks
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Scuba diving
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Respiration in high altitude conditions
Raoult’s Law
The partial vapor pressure of each component of a solution is directly proportional to its mole fraction.
pA=pA0⋅xAp_A = p_A^0 \cdot x_A
For a solution:
ptotal=pA+pBp_{total} = p_A + p_B
where pA0p_A^0 and pB0p_B^0 are vapor pressures of pure components.
Ideal and Non-Ideal Solutions
Ideal Solutions
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Obey Raoult’s Law over the entire concentration range.
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No enthalpy change (ΔHmix = 0).
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No volume change (ΔVmix = 0).
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Example: Benzene + Toluene, n-hexane + n-heptane.
Non-Ideal Solutions
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Deviate from Raoult’s Law.
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Show positive or negative deviations.
Positive Deviation:
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Weaker intermolecular interactions.
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ΔHmix > 0, ΔVmix > 0.
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Example: Ethanol + Acetone.
Negative Deviation:
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Stronger intermolecular interactions.
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ΔHmix < 0, ΔVmix < 0.
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Example: Chloroform + Acetone.
Colligative Properties
Colligative properties depend only on the number of solute particles, not their nature.
1. Relative Lowering of Vapor Pressure
Δpp0=χsolute\frac{\Delta p}{p^0} = \chi_{solute}
2. Elevation of Boiling Point
ΔTb=Kb⋅m\Delta T_b = K_b \cdot m
where KbK_b = ebullioscopic constant, mm = molality.
3. Depression of Freezing Point
ΔTf=Kf⋅m\Delta T_f = K_f \cdot m
where KfK_f = cryoscopic constant.
4. Osmotic Pressure
π=CRT\pi = C R T
where CC = molar concentration, RR = gas constant, TT = temperature in Kelvin.
Abnormal Molar Mass
Sometimes experimental molar mass differs from theoretical molar mass due to association or dissociation of solute particles.
The van’t Hoff factor (i) is used to correct colligative properties.
i=Observed colligative propertyCalculated colligative propertyi = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}
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Dissociation (i > 1) → e.g., NaCl in water.
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Association (i < 1) → e.g., acetic acid in benzene.
Important Applications of Solutions
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Preservation of food using salt and sugar.
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Determination of molar masses using colligative properties.
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Industrial applications in antifreeze, pharmaceuticals, beverages.
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Medical uses like IV fluids, osmotic pressure-based treatments.
Key Notes for Revision
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Solution = Solute + Solvent.
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Concentration terms: Molarity, Molality, Mole fraction, ppm.
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Solubility of gases follows Henry’s law.
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Raoult’s law explains vapor pressure.
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Colligative properties: lowering of vapor pressure, elevation of boiling point, depression of freezing point, osmotic pressure.
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Van’t Hoff factor accounts for abnormal molar mass.
Multiple Choice Questions (MCQs)
Class 12 Chemistry Solutions Notes
Q1. Which of the following is an example of a solid solution?
a) Oxygen in water
b) Brass
c) Soda water
d) Alcohol in water
Answer: Brass
Q2. The solute present in lesser amount in a solution is called:
a) Solvent
b) Solute
c) Mixture
d) Alloy
Answer: Solute
Q3. The concentration unit independent of temperature is:
a) Molarity
b) Normality
c) Molality
d) Mass by volume
Answer: Molality
Q4. The solubility of gases in liquids increases with:
a) Increase in temperature
b) Decrease in pressure
c) Increase in pressure
d) Decrease in solute
Answer: Increase in pressure
Q5. Henry’s law is related to:
a) Freezing point
b) Solubility of gas in liquid
c) Osmotic pressure
d) Boiling point
Answer: Solubility of gas in liquid
Q6. Which of the following is an ideal solution?
a) Ethanol + Acetone
b) Benzene + Toluene
c) Chloroform + Acetone
d) HCl + Water
Answer: Benzene + Toluene
Q7. Negative deviation from Raoult’s law is shown by:
a) Alcohol + Water
b) Acetone + Ethanol
c) Chloroform + Acetone
d) Benzene + Toluene
Answer: Chloroform + Acetone
Q8. The vapor pressure of pure solvent is 0.8 atm. After adding a non-volatile solute, it becomes 0.6 atm. The mole fraction of solute is:
a) 0.25
b) 0.5
c) 0.75
d) 0.4
Answer: 0.25
Q9. Colligative properties depend upon:
a) Nature of solute
b) Number of solute particles
c) Type of solvent
d) Molar mass of solvent
Answer: Number of solute particles
Q10. Which one is not a colligative property?
a) Osmotic pressure
b) Elevation of boiling point
c) Depression of freezing point
d) Viscosity
Answer: Viscosity
Q11. The cryoscopic constant is related to:
a) Freezing point depression
b) Boiling point elevation
c) Osmotic pressure
d) Vapor pressure
Answer: Freezing point depression
Q12. Osmotic pressure is represented by which formula?
a) π = C R T
b) π = m Kf
c) π = ΔTb / Kb
d) π = p°χ
Answer: π = C R T
Q13. When NaCl dissolves in water, the van’t Hoff factor (i) is:
a) 0
b) 1
c) 2
d) 3
Answer: 2
Q14. Which solution shows association of solute molecules?
a) NaCl in water
b) Acetic acid in benzene
c) KCl in water
d) HCl in water
Answer: Acetic acid in benzene
Q15. The unit of molarity is:
a) mol/kg
b) mol/L
c) g/L
d) g/kg
Answer: mol/L
Q16. Depression in freezing point is directly proportional to:
a) Molality
b) Molarity
c) Normality
d) Temperature
Answer: Molality
Q17. Which of the following is used for preserving food items?
a) Osmotic pressure
b) Colligative property
c) High concentration of salt/sugar
d) Raoult’s law
Answer: High concentration of salt/sugar
Q18. The vapor pressure of a liquid decreases when:
a) Solute is added
b) Pressure is increased
c) Temperature is increased
d) Volume is decreased
Answer: Solute is added
Q19. The boiling point of a solution is always:
a) Lower than pure solvent
b) Higher than pure solvent
c) Same as pure solvent
d) Cannot be determined
Answer: Higher than pure solvent
Q20. Van’t Hoff factor (i) is equal to 1 for:
a) NaCl in water
b) K2SO4 in water
c) Glucose in water
d) CH3COOH in benzene
Answer: Glucose in water
Short Answer Questions-Class 12 Chemistry Solutions Notes
Q1. Define molality and write its formula.
Answer: Molality is defined as the number of moles of solute dissolved in 1 kilogram of solvent. Its formula is:
m=Moles of soluteMass of solvent in kgm = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}
Q2. State Raoult’s law for a solution containing a non-volatile solute.
Answer: Raoult’s law states that the relative lowering of vapor pressure of a solvent containing a non-volatile solute is equal to the mole fraction of solute present in the solution.
Q3. Why is molality preferred over molarity in expressing concentration?
Answer: Molality is preferred because it depends on the mass of the solvent, which does not change with temperature, while molarity depends on the volume of solution, which changes with temperature.
Q4. What is meant by positive deviation from Raoult’s law?
Answer: Positive deviation occurs when the intermolecular forces between solute and solvent are weaker than those in pure components, resulting in higher vapor pressure than expected.
Q5. Name the colligative property used to determine the molar mass of macromolecules.
Answer: Osmotic pressure is used to determine the molar mass of macromolecules like proteins and polymers.
Q6. Define cryoscopic constant.
Answer: Cryoscopic constant (Kf) is the depression in freezing point of a solvent when 1 mole of solute is dissolved in 1 kilogram of solvent.
Q7. What is the effect of pressure on the solubility of solids in liquids?
Answer: The solubility of solids in liquids is practically unaffected by pressure because solids and liquids are incompressible.
Q8. Write the mathematical expression of Henry’s law.
Answer: Henry’s law can be expressed as:
p=kH⋅xp = k_H \cdot x
where pp is the partial pressure of the gas, xx is the mole fraction of the gas in solution, and kHk_H is Henry’s law constant.
Q9. What is abnormal molar mass?
Answer: When the experimentally determined molar mass of a solute using colligative properties differs from the theoretical molar mass, it is called abnormal molar mass.
Q10. What is van’t Hoff factor (i)?
Answer: Van’t Hoff factor (i) is the ratio of the observed colligative property to the calculated colligative property. It accounts for association or dissociation of solute particles.
Long Answer Questions (LAQs)-Class 12 Chemistry Solutions Notes
Q1. Explain the factors affecting solubility of a solid in a liquid.
Answer: The solubility of a solid in a liquid depends mainly on:
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Nature of solute and solvent: Polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents.
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Temperature: For most solids, solubility increases with temperature due to endothermic dissolution.
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Pressure: Pressure has negligible effect because solids and liquids are nearly incompressible.
Q2. Derive the expression for elevation of boiling point.
Answer:
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When a non-volatile solute is added to a solvent, its vapor pressure decreases.
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As a result, a higher temperature is required to make vapor pressure equal to external pressure.
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The elevation in boiling point is proportional to the molality of the solution:
ΔTb=Kb⋅m\Delta T_b = K_b \cdot m
where KbK_b is the ebullioscopic constant and mm is molality.
Q3. Differentiate between ideal and non-ideal solutions with examples.
Answer:
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Ideal Solutions: Obey Raoult’s law at all concentrations, ΔHmix = 0, ΔVmix = 0. Example: Benzene + Toluene.
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Non-Ideal Solutions: Deviate from Raoult’s law. They show either positive deviation (e.g., Ethanol + Acetone) or negative deviation (e.g., Chloroform + Acetone).
Q4. What are colligative properties? Explain any two in detail.
Answer: Colligative properties are properties of solutions that depend only on the number of solute particles and not on their nature. Examples include:
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Relative lowering of vapor pressure: Addition of solute lowers the vapor pressure of solvent, proportional to mole fraction of solute.
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Depression of freezing point: A solution freezes at a lower temperature than pure solvent due to disruption of crystal formation by solute particles.
Q5. Discuss the applications of Henry’s law in daily life.
Answer: Applications of Henry’s law include:
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Carbonated beverages: CO₂ is dissolved under high pressure to maintain fizz.
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Scuba diving: At high pressure, nitrogen dissolves in blood. Rapid ascent causes bubbles leading to bends.
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High altitude breathing: Low atmospheric pressure reduces O₂ solubility in blood, causing altitude sickness.
Conclusion
The chapter Solutions builds a strong conceptual foundation about mixtures, their concentration, properties, and real-life significance. Understanding formulas and laws like Raoult’s law, Henry’s law, and colligative properties is crucial for both theoretical and application-based problems. With clear practice, this chapter becomes scoring in Class 12 exams as well as in competitive examinations.
Solid State Class 12 Chemistry Notes | Complete Chapter, MCQs, Short & Long Q&A