Solutions Chapter Notes | CBSE Class 12 Chemistry

Key Terms and Definitions

Solution

A homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

Molarity (M)

Number of moles of solute dissolved per litre of the solution. Unit: mol/L. It is temperature-dependent.

Molality (m)

Number of moles of solute dissolved per kilogram of the solvent. Unit: mol/kg. It is temperature-independent.

Mole Fraction (χ)

Ratio of the number of moles of one component to the total number of moles of all components in the solution. It is a dimensionless quantity.

Henry's Law

The partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (χ) in the solution. Formula: p = Kʜ * χ.

Raoult's Law

For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

Colligative Properties

Properties of a solution that depend on the number of solute particles present, but not on their identity or nature.

Azeotrope

A binary liquid mixture that has a constant boiling point and a composition that does not change upon distillation.

Van't Hoff Factor (i)

The ratio of the observed colligative property to the calculated colligative property. It accounts for the extent of dissociation or association of solute particles in a solution.

Concentration Terms: Formula Sheet

• Molarity (M): M = (Moles of Solute) / (Volume of Solution in L)
• Molality (m): m = (Moles of Solute) / (Mass of Solvent in kg)
• Mole Fraction (χ): For component A, χ_A = (Moles of A) / (Total Moles of all components)
• Mass Percentage (% w/w): % w/w = [(Mass of Solute) / (Mass of Solution)] * 100
• Volume Percentage (% v/v): % v/v = [(Volume of Solute) / (Volume of Solution)] * 100
• Parts Per Million (ppm): ppm = [(Mass of Solute) / (Mass of Solution)] * 10^6
• Relationship: Molarity is temperature-dependent because volume changes with temperature. Molality is temperature-independent as it is based on mass.

Raoult's Law, Ideal and Non-Ideal Solutions

Raoult's Law is a cornerstone of this chapter. For a solution containing two volatile components A and B, the law states that the partial vapour pressure of each component (P_A, P_B) is the product of the vapour pressure of the pure component (P°_A, P°_B) and its mole fraction (χ_A, χ_B) in the solution.
**P_A = P°_A χ_A and P_B = P°_B χ_B**.
According to Dalton's law of partial pressures, the total pressure P_total is **P_total = P_A + P_B = P°_A χ_A + P°_B χ_B**.

Solutions are classified based on their adherence to this law:

• Ideal Solutions: These solutions obey Raoult's law perfectly over the entire range of concentration. This happens when the intermolecular forces between solute-solvent molecules (A-B) are nearly identical to those between solute-solute (A-A) and solvent-solvent (B-B) molecules. For ideal solutions, the enthalpy of mixing (ΔH_mix = 0) and the volume of mixing (ΔV_mix = 0). Examples include n-hexane and n-heptane, or bromoethane and chloroethane.

• Non-Ideal Solutions: These do not obey Raoult's law and show deviations.
• Positive Deviation: The observed vapour pressure is higher than predicted by Raoult's law (P_total > P°_Aχ_A + P°_Bχ_B). This occurs when A-B interactions are weaker than A-A and B-B interactions. Mixing is endothermic (ΔH_mix > 0) and results in a volume increase (ΔV_mix > 0). Example: Ethanol and acetone. These form minimum boiling azeotropes.
• Negative Deviation: The observed vapour pressure is lower than predicted (P_total < P°_Aχ_A + P°_Bχ_B). This occurs when A-B interactions are stronger than A-A and B-B interactions, often due to new hydrogen bond formation. Mixing is exothermic (ΔH_mix < 0) and results in a volume decrease (ΔV_mix < 0). Example: Chloroform and acetone. These form maximum boiling azeotropes.

Ideal vs. Non-Ideal Solutions: Quick Comparison

Colligative Properties: Must-Remember Formulas

• Definition: Properties that depend only on the number of solute particles relative to the total number of particles, not on the nature of the solute.
• 1. Relative Lowering of Vapour Pressure (RLVP): (P°_A - P_A) / P°_A = χ_B (where A is solvent, B is non-volatile solute).
• 2. Elevation in Boiling Point (ΔT_b): ΔT_b = K_b * m. Where K_b is the Ebullioscopic Constant.
• 3. Depression in Freezing Point (ΔT_f): ΔT_f = K_f * m. Where K_f is the Cryoscopic Constant.
• 4. Osmotic Pressure (π): π = CRT = (n/V)RT. Where C is molar concentration, R is the gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin.
• For Electrolytes (Abnormal Molar Mass): All colligative property formulas must be modified by the Van't Hoff factor (i).
• Modified Formula for Elevation in Boiling Point: ΔT_b = i K_b m
• Modified Formula for Depression in Freezing Point: ΔT_f = i K_f m
• Modified Formula for Osmotic Pressure: π = i C R * T

Calculating Van't Hoff Factor (i)

• {"header":"Concept:","body":"The Van't Hoff factor (i) corrects for the dissociation or association of solutes. i = (Normal Molar Mass) / (Abnormal Molar Mass) = (Observed Colligative Property) / (Calculated Colligative Property). For non-electrolytes like sugar or urea, i = 1."}
• {"header":"Example 1: Strong Electrolyte Dissociation","body":"For 100% dissociation of K₂SO₄:\nK₂SO₄ → 2K⁺ + SO₄²⁻\nOne formula unit produces 3 ions (2 K⁺ and 1 SO₄²⁻). Therefore, i = 3."}
• {"header":"Example 2: Association","body":"Acetic acid (CH₃COOH) dimerizes in benzene:\n2CH₃COOH ⇌ (CH₃COOH)₂\nTwo molecules associate into one. For 100% association, i = 1/2 = 0.5."}
• {"header":"Example 3: Worked Problem","body":"Calculate the osmotic pressure of a 0.01 M MgCl₂ solution at 27°C (300 K), assuming complete dissociation. (R=0.0821 L atm/mol K)\nStep 1: Find 'i'. MgCl₂ → Mg²⁺ + 2Cl⁻. Number of ions = 1 + 2 = 3. So, i = 3.\nStep 2: Use the modified formula. π = iCRT\nπ = 3 (0.01 mol/L) (0.0821 L atm/mol K) * (300 K)\nπ = 0.7389 atm"}

Board Exam Trap Alert

Trap 1: Forgetting the Van't Hoff Factor (i). This is the most common mistake in numericals. Before solving any colligative property problem, ask yourself: 'Is the solute an electrolyte (acid, base, salt)?' If yes, you MUST calculate and use 'i'. Forgetting it for NaCl, K₂SO₄, or CaCl₂ will lead to incorrect answers and significant mark loss.

Trap 2: Molarity vs. Molality Units. Be careful with units. Molality (m) is moles/kg of solvent. Molarity (M) is moles/L of solution. When given density, you might need to convert between them. Always write the formula and check your units before calculating.

Quick Revision Check

• Why do deep-sea divers use breathing air diluted with helium? To avoid 'the bends'. At high pressure underwater, nitrogen from air dissolves in the blood. When the diver ascends, pressure decreases, and dissolved nitrogen forms bubbles in the blood, which is painful and dangerous. Helium is much less soluble in blood than nitrogen, preventing this condition.
• What is reverse osmosis? It is the process of moving a solvent from a region of high solute concentration to a region of low solute concentration through a semi-permeable membrane by applying an external pressure greater than the osmotic pressure. It is used in the desalination of seawater.
• Out of 1 M glucose and 1 M NaCl, which solution will have a higher boiling point and why? 1 M NaCl will have a higher boiling point. Elevation in boiling point is a colligative property. NaCl is an electrolyte that dissociates into two ions (i=2), while glucose is a non-electrolyte (i=1). NaCl produces more particles in solution, leading to a greater elevation in boiling point.
• What is an ebullioscopic constant (Kb)? It is the elevation in boiling point observed when the molality of a solution is unity (1 mol/kg). Its unit is K kg/mol.

Frequently Asked Questions

Frequently Asked Questions