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Chapter 2 — Solutions

Class 12 · Chemistry

Overview

Chapter 2 — Solutions Master Diagram

Introduction: "Solutions" is a core chapter in Class 12 Chemistry (Chemistry — Part I) that develops the physical chemistry concepts needed to understand how substances mix and how those mixtures behave. It introduces concentration terms, the thermodynamic basis of vapour pressure of mixtures, and the phenomena collectively called colligative properties. The chapter connects microscopic ideas (mole fraction, solute–solvent interactions, dissociation/association) with measurable macroscopic effects (vapour pressure change, boiling/freezing point shifts, osmotic pressure). Importance: Mastery of this chapter is essential for problem solving in physical chemistry and for practical applications in industry, biology and everyday life (e.g., antifreeze, desalination, determination of molar masses). It is also frequently tested in board exams through derivations, concept-based questions and numerical problems. Key themes: The chapter covers concentration units (mole fraction, molality, molarity, mass percent, ppm), Raoult’s law for ideal solutions and deviations from it (positive and negative deviations, azeotropes), Henry’s law for gases in liquids, and the four main colligative…

Learning Objectives

  • Define solution, solute, solvent and classify solutions based on physical state and concentration.
  • Explain mole fraction, molality, molarity and mass percent and convert between these concentration units.
  • Calculate composition of solutions in terms of mole fraction, molality, molarity and percent by mass from given data.
  • Derive and apply Raoult's law for ideal solutions and explain positive and negative deviations from Raoult's law.
  • Explain vapour pressure lowering for solutions containing non-volatile solutes and calculate ΔP using Raoult's law.
  • Derive and apply the equations for boiling point elevation and freezing point depression and compute ΔTb and ΔTf.
  • Explain osmotic pressure and use the relation π = MRT to calculate osmotic pressure and determine molar mass of solutes.
  • Determine molar mass of polymers and low-molecular-weight solutes using colligative property measurements (ΔTb, ΔTf, π).

Topics in this chapter

11 topics · tap a topic title to jump straight to it.

🧴1

Introduction to Solutions

Fig 1 — Educational Diagram: Introduction to Solutions

Fig 1 — Educational Diagram: Introduction to Solutions

⚗️ CHEMICAL REACTION

Introduction to Solutions

Core Principle: Molarity: M = n_solute (mol) / V_solution (L)

What is a solution?
A solution is a homogeneous mixture of two or more substances. The component present in larger amount is called the solvent (usually a liquid) and the component(s) present in smaller amount are called solute(s). Examples: salt in water, sugar in tea, alloys like brass.

Types of solutions
- By physical state: solid in liquid (NaCl in water), liquid in liquid (ethanol in water), gas in liquid (CO2 in soda), solid in solid (alloys).
- By composition: binary (two components), ternary (three components).
- Based on saturation: unsaturated (can dissolve more solute), saturated (equilibrium between dissolving and crystallizing), supersaturated (contains more solute than at equilibrium; metastable).

Concentration measures
Different ways to express how much solute is present in a solution (useful depending on context):

  • Molarity (M) — moles of solute per litre of solution: M = n_solute (mol) / V_solution (L).
  • Molality (m) — moles of solute per kilogram of solvent: m = n_solute (mol) / mass_solvent (kg). Useful when temperature changes because it does not depend on volume.
  • Mole fraction (x) — ratio of moles of a component to total moles: x_i = n_i / Σ n_j (dimensionless).
  • Mass percent (w/w %) = (mass_solute / mass_solution) × 100%.
  • Volume percent (v/v %) = (volume_solute / volume_solution) × 100% (used for miscible liquids).
  • ppm / ppb — parts per million/ billion for very dilute solutions: ppm = (mass_solute / mass_solution) × 10^6.

Ideal and real solutions
An ideal solution obeys Raoult's law for all compositions; interactions between like and unlike molecules are similar. Most real solutions deviate from ideality — positive deviation when solvent–solute interactions are weaker, negative when stronger.

Key laws (brief)
- Raoult's law (for an ideal solution): P_i = x_i · P_i° where P_i is partial vapour pressure of component i, x_i its mole fraction in liquid phase and P_i° the vapour pressure of pure i.
- Henry's law (for gases dissolved in liquids at low concentration): C = k_H · p (concentration of dissolved gas ∝ partial pressure above the liquid).

Factors affecting solubility
Nature of solute & solvent ("like dissolves like"—polar dissolves polar), temperature (solubility of most solids in water increases with temperature; solubility of gases decreases with temperature), pressure (significant for gases: increased pressure increases gas solubility).

Why concentrations matter
Concentration units are used to calculate reaction rates, equilibria, colligative properties (boiling point elevation, freezing point depression, osmotic pressure), and to design practical solutions (pharmaceutical, industrial, laboratory).

📌 Examples
  • Table salt (NaCl) dissolved in water — a typical solid-in-liquid solution; used in cooking and saline preparation.
  • Sugar in tea or coffee — concentration commonly expressed in g per cup or molarity for experiments.
  • Carbonated beverages — CO2 dissolved in water under pressure; on opening pressure drops and gas escapes (Henry's law).
  • Brass (alloy of copper and zinc) — solid solution used in coins and instruments.
  • Antifreeze — ethylene glycol in water lowers freezing point (colligative effect), used in car radiators.
🧮 Formulas
  1. \[Molarity: M = n_solute (mol) / V_solution (L)\]
  2. \[Molality: m = n_solute (mol) / mass_of_solvent (kg)\]
  3. \[Mole fraction: x_i = n_i / Σ n_j\]
  4. \[Mass percent: w/w % = (mass_solute / mass_solution) × 100\]
  5. \[Volume percent: v/v % = (volume_solute / volume_solution) × 100\]
  6. \[ppm: = (mass_solute / mass_solution) × 10^6\]
⚖️2

Concentration Terms and Expressions

Fig 2 — Educational Diagram: Concentration Terms and Expressions

Fig 2 — Educational Diagram: Concentration Terms and Expressions

⚗️ CHEMICAL REACTION

Concentration Terms and Expressions

Core Principle: Molarity: M = n_solute (mol) / V_solution (L)

Overview
Concentration terms quantify how much solute is present in a given quantity of solution or solvent. Different expressions are used depending on experimental convenience and on whether the effect of the solute depends on amount of solute per mass of solvent (colligative properties) or per volume of solution (reactivity, stoichiometry).

Common concentration terms (definitions)

  • Molarity (M): moles of solute per litre of solution. M = moles solute / L solution.
  • Molality (m): moles of solute per kilogram of solvent. m = moles solute / kg solvent. Independent of temperature.
  • Mole fraction (x_i): ratio of moles of component i to total moles. x_i = n_i / (Σ n_j). Dimensionless.
  • Mass percent (w/w %): (mass solute / mass solution) × 100%.
  • Volume percent (v/v %): (volume solute / volume solution) × 100% (used for miscible liquids).
  • Mass/volume percent (w/v %): (mass solute (g) / volume solution (mL)) × 100% (common in biology/medicine).
  • Parts per million (ppm) and parts per billion (ppb): ppm = (mass solute / mass solution) × 10^6; ppb = ×10^9. Used for very dilute contaminants.
  • Normality (N): equivalents of solute per litre of solution. N = equivalents / L solution. For acid–base reactions equivalents = moles × basicity or acidity (or charge for redox).
  • Formality (F): molar concentration expressed in terms of formula units per litre of solution (used when species dissociate; initial analytical concentration before dissociation).

Why choose one expression over another?
Use molality for colligative properties (ΔTb, ΔTf, osmotic pressure per mole of solute, independent of T). Use molarity for reaction stoichiometry and volumetric titrations. Use mole fraction for Raoult's law and vapour-pressure relationships. Use ppm/ppb for trace analysis and environmental measurements.

Key practical points
- Molarity depends on temperature because volume changes with temperature. Molality does not.
- For electrolyte solutions, account for dissociation using van't Hoff factor i (effective number of particles). For colligative effects, replace m by i · m (or use effective concentration).

Conversions — important relations (idea)
For 1 L of solution of molarity M and solution density d (g mL^-1):

  • Mass of solute = M · M_r (g) (where M_r is molar mass).
  • Mass of solution = 1000 · d (g).
  • Mass of solvent = 1000·d − M·M_r (g) ⇒ in kg: (1000·d − M·M_r)/1000.
  • Molality m = M·1000 / (1000·d − M·M_r).
  • If solvent is water (M_w = 18.015 g mol^-1), mole fraction of solute x_solute = M / (M + (mass solvent in g)/18.015).

Colligative relations (use molality)

  • Freezing point depression: ΔTf = Kf · m · i
  • Boiling point elevation: ΔTb = Kb · m · i
  • Osmotic pressure: π = i · M · R · T (for dilute solutions; M is molarity)

Practical tips for lab work
When preparing solutions, decide whether concentration should be expressed per litre of solution (molar) or per mass of solvent (molal). For volumetric analysis and titrations use molarity or normality (N = M × n where n is number of H+ or OH- or electrons per formula unit for the reaction). For measuring colligative properties use molality.

📌 Examples
  • Preparing 1.0 L of 0.1 M NaCl: molar mass NaCl = 58.44 g mol^-1. Mass needed = 0.1 mol × 58.44 g mol^-1 = 5.844 g. Dissolve in water and make the volume up to 1.0 L.
  • Convert 0.5 M glucose (C6H12O6, M_r = 180.16 g mol^-1) to molality assuming solution density ≈ 1.00 g mL^-1: For 1 L, mass solute = 0.5×180.16 = 90.08 g; mass solvent ≈ 1000 − 90.08 = 909.92 g = 0.90992 kg. Molality m = 0.5 mol / 0.90992 kg ≈ 0.549 m.
  • ppm example: A water sample contains 2 mg of lead per kg of water. Lead concentration = 2 ppm (since ppm = mg solute/kg solution).
  • Colligative property example: 0.10 m (molal) solution of a non-electrolyte has freezing-point depression ΔTf = Kf · m. For water Kf = 1.86 °C kg mol^-1, so ΔTf = 1.86×0.10 = 0.186 °C (freezing point depressed by 0.186 °C).
  • Normality example (acid titration): 0.1 M H2SO4 (diprotic) has N = M × equivalents per mole = 0.1 × 2 = 0.2 N for acid–base neutralisation.
🧮 Formulas
  1. \[Molarity: M = n_solute (mol) / V_solution (L)\]
  2. \[Molality: m = n_solute (mol) / mass_solvent (kg)\]
  3. \[Mole fraction: x_i = n_i / Σ n_j\]
  4. \[Mass percent: w/w % = (mass_solute / mass_solution) × 100\]
  5. \[ppm = (mass_solute / mass_solution) × 10^6\]
    \[ppb = × 10^9\]
  6. \[Normality: N = equivalents of solute / V_solution (L) = M × (equivalents per mole)\]
🎈3

Vapour Pressure of Solutions and Raoult's Law

Fig 3 — Educational Diagram: Vapour Pressure of Solutions and Raoult

Fig 3 — Educational Diagram: Vapour Pressure of Solutions and Raoult's Law

📜 THEOREM / LAW

Vapour Pressure of Solutions and Raoult's Law

Core Principle: p_A = x_A · p_A^0

Vapour pressure is the pressure exerted by the vapour in dynamic equilibrium with its liquid at a given temperature. When a non-volatile or volatile solute is mixed with a solvent, the vapour pressure of the solvent over the solution usually changes.

Raoult's law (ideal solutions)
For an ideal solution, the partial vapour pressure of each volatile component is directly proportional to its mole fraction in the liquid phase. For component A:

pA = xA · pA0

where pA is the partial vapour pressure of A over the solution, xA is the mole fraction of A in the liquid, and pA0 is the vapour pressure of pure A at the same temperature.

Binary mixture (A + B)
Total vapour pressure: P = pA + pB = xApA0 + xBpB0 (with xA + xB = 1).

Non-volatile solute
If a non-volatile solute is dissolved in a volatile solvent, only the solvent contributes to vapour. Raoult's law gives:

p = xsolvent · psolvent0

Vapour pressure lowering: Δp = psolvent0 − p = psolvent0 · xsolute. Relative lowering: Δp / psolvent0 = xsolute.

Meaning and consequences
- Vapour pressure lowering is a colligative property: at a given temperature it depends on the number of solute particles (mole fraction) and not on their identity (for non-electrolytes and ideal behaviour).
- Lower vapour pressure of a solution means higher boiling point (boiling point elevation) and lower freezing point (freezing point depression).

Ideal vs non-ideal behaviour
- Ideal solution: A–A, B–B and A–B interactions are similar → Raoult's law holds over entire composition range.
- Positive deviation: A–B interactions are weaker than A–A and B–B → observed vapour pressure > Raoult's prediction → lower boiling point; may give a minimum-boiling azeotrope (e.g., ethanol–water azeotrope).
- Negative deviation: A–B interactions stronger than A–A and B–B → observed vapour pressure < Raoult's prediction → higher boiling point; may give a maximum-boiling azeotrope (e.g., chloroform–acetone shows strong specific interaction).

When is Raoult's law applicable?
- Solutions of chemically similar substances (e.g., benzene–toluene), or at low solute concentrations for solute acting as ideal dilute solute. For dilute non-volatile solute in a solvent, the solvent follows Raoult's law while solute follows Henry's law at very low concentrations.

How to use: Measure or use known vapour pressures of pure components (p0) and mole fractions to calculate partial and total vapour pressures and predict boiling behaviour or compositions of vapor in distillation problems.

📌 Examples
  • Adding table salt (non-volatile) to water lowers the vapour pressure of water — this makes salted water boil at a slightly higher temperature than pure water (boiling point elevation).
  • Ethanol–water mixture shows deviation from Raoult's law and forms a boiling azeotrope (≈95.6% ethanol) used in distillation limitations during alcohol purification.
  • Sugar in fruit preserves reduces the effective vapour pressure of water, helping reduce microbial activity and slow spoilage (colligative effect).
  • Binary ideal-like mixture: benzene and toluene approximately obey Raoult's law; partial pressures of each vary linearly with mole fraction.
🧮 Formulas
  1. \[p_A = x_A · p_A^0\]
  2. \[p_B = x_B · p_B^0\]
  3. \[Total pressure: P = p_A + p_B = x_A p_A^0 + x_B p_B^0\]
  4. \[For non-volatile solute: p = x_solvent · p_solvent^0\]
  5. \[Vapour pressure lowering: Δp = p^0 - p = p^0 · x_solute\]
  6. \[Relative lowering: Δp / p^0 = x_solute\]
🔬4

Colligative Properties — Overview

Fig 4 — Educational Diagram: Colligative Properties — Overview

Fig 4 — Educational Diagram: Colligative Properties — Overview

⚗️ CHEMICAL REACTION

Colligative Properties — Overview

Core Principle: Raoult's law (non-volatile solute): P_solvent = X_solvent · P°_solvent, where X_solvent is mole fraction of solvent and P°_solvent is vapour pressure of pure solvent.

Definition: Colligative properties are physical properties of solutions that depend only on the number (concentration) of solute particles present, and not on their chemical identity (for ideally behaving, non-electrolyte solutes).

Four main colligative properties:

  • Vapor pressure lowering
  • Boiling point elevation
  • Freezing point depression
  • Osmotic pressure

How they arise (qualitative idea): Adding a non-volatile solute to a solvent lowers the chemical potential of the solvent in the liquid phase. As a result the solvent's vapour pressure above the solution is lower than that of the pure solvent (Raoult's law). Because vapour pressure is reduced, a higher temperature is needed to reach the external pressure (boiling point elevation) and a lower temperature is needed for solidification (freezing point depression). Lowered chemical potential also leads to osmotic flow (solvent moves from pure solvent to solution).

Important assumptions & corrections: Formulas below are strictly for dilute, ideally behaving solutions. For ionic solutes (electrolytes) dissociation produces more particles; the observed effect is multiplied by the van't Hoff factor i (effective number of particles per formula unit). At higher concentrations deviations occur due to ion pairing and non-ideal interactions (use activities).

Practical uses: Determination of molar mass of solutes (especially polymers using osmotic pressure), antifreeze formulations, de-icing roads (freezing point depression), reverse osmosis and water purification, biological osmotic balance (cells, plasmolysis).

📌 Examples
  • Sprinkling salt on icy roads: salt lowers the freezing point of water so ice melts at temperatures below 0°C (freezing point depression).
  • Antifreeze (ethylene glycol) in car radiators: raises boiling point and lowers freezing point of coolant, preventing boil-over and freezing.
  • Seawater boils at a slightly higher temperature than pure water because dissolved salts raise the boiling point.
  • Sugar or salt in jam/honey creates high osmotic pressure that inhibits microbial growth (preservation).
  • Reverse osmosis: applying pressure greater than the osmotic pressure forces water through a membrane, desalinating seawater.
  • Polymer molar mass measurement: using osmotic pressure of dilute polymer solutions to determine high molar masses.
🧮 Formulas
  1. \[Raoult's law (non-volatile solute): P_solvent = X_solvent · P°_solvent\]
    \[where X_solvent is mole fraction of solvent and P°_solvent is vapour pressure of pure solvent.\]
  2. \[Vapour pressure lowering: ΔP = P°_solvent - P_solvent = X_solute · P°_solvent (approximately for dilute solutions).\]
  3. \[Boiling point elevation: ΔT_b = K_b · m · i\]
    \[where K_b is the ebullioscopic constant\]
    \[m is molality (mol kg⁻¹)\]
    \[and i is the van't Hoff factor (i = 1 for non-electrolytes).\]
  4. \[Freezing point depression: ΔT_f = K_f · m · i\]
    \[where K_f is the cryoscopic constant.\]
  5. \[Osmotic pressure (ideal dilute solution): π = i · M · R · T\]
    \[where M is molarity (mol L⁻¹)\]
    \[R = 0.08206 L·atm·K⁻¹·mol⁻¹ (or appropriate units)\]
    \[T is absolute temperature (K).\]
  6. \[Molar mass from osmotic pressure: M = (w · R · T) / (π · V)\]
    \[where w = mass of solute (g)\]
    \[V = volume of solution (L), π in same pressure units as R\]
    \[and M in g·mol⁻¹.\]
🔬5

Boiling Point Elevation and Freezing Point Depression

Fig 5 — Educational Diagram: Boiling Point Elevation and Freezing Point Depression

Fig 5 — Educational Diagram: Boiling Point Elevation and Freezing Point Depression

⚗️ CHEMICAL REACTION

Boiling Point Elevation and Freezing Point Depression

Core Principle: Raoult's law (for solvent in ideal solution): p_solvent = x_solvent · p°_solvent

Overview (colligative properties): Boiling point elevation and freezing point depression are colligative properties — they depend only on the number of solute particles dissolved in a solvent, not on their chemical identity. When a non-volatile solute is added to a solvent, the vapor pressure of the solvent is lowered. This causes the boiling point to rise (because a higher temperature is needed for the vapor pressure to reach the external pressure) and the freezing point to fall (because the equilibrium between solid and liquid phases is shifted).

Physical basis:

  • Vapor-pressure lowering (Raoult's law): For an ideal solution with a nonvolatile solute, p_solvent = x_solvent · p°_solvent. Adding solute reduces x_solvent and therefore the vapor pressure.
  • Boiling point elevation: At constant external pressure, the temperature at which vapor pressure equals the external pressure increases when vapor pressure is reduced by solute.
  • Freezing point depression: The chemical potential of the liquid solvent is lowered by the solute, so equilibrium between liquid and solid shifts to lower temperature — the solution must be cooled further to freeze.

Key assumptions and limits: Results below apply to dilute, ideally behaving solutions. For electrolytes, the effective number of particles is multiplied by the van't Hoff factor (i), but ion pairing and non-ideal behavior at higher concentrations reduce i from its ideal integer value.

Derivation outline (sketch):

  • Start with Raoult's law for solvent vapor pressure in a dilute solution: p = x_solvent · p°.
  • For small solute mole fraction x_solute, x_solvent ≈ 1 - x_solute, so vapor-pressure lowering Δp ≈ x_solute · p°.
  • Use Clausius-Clapeyron / Clapeyron relations to relate change in vapor pressure to change in temperature and the enthalpy of vaporization (or fusion). Linearization for small ΔT yields ΔTb proportional to solute molality m.
  • Result: ΔTb = Kb · m (and ΔTf = Kf · m) for non-electrolytes; for electrolytes multiply by the van't Hoff factor i: ΔT = i · K · m.

Practical numbers for water: Kb (water) = 0.512 K kg mol⁻¹; Kf (water) = 1.86 K kg mol⁻¹. These are experimentally tabulated values.

Important consequences and notes:

  • Boiling point elevation and freezing point depression are proportional to molality (moles solute per kg solvent), so they are independent of temperature and volume effects associated with concentration expressed as molarity.
  • Electrolytes produce more particles; for example, ideally NaCl gives i ≈ 2, CaCl2 gives i ≈ 3. Real solutions often show i < ideal because of ion pairing.
  • These effects are small for typical household concentrations. For example, adding 1 molal NaCl to water raises its boiling point by about 1.02 K and lowers its freezing point by about 3.72 K (using ideal i = 2).

Applications: De-icing roads, automotive antifreeze, ice-cream making, osmometry and molar mass determinations, preserving cells/tissues (cryobiology) and many environmental impacts (seawater freezing point is lower than pure water).

End of summary for Class 12 CBSE level.

📌 Examples
  • De-icing roads: Spreading NaCl on icy roads lowers the freezing point of water so ice melts at sub-zero temperatures (practical until the salt solution becomes too cold for effectiveness).
  • Automotive antifreeze: Ethylene glycol (or propylene glycol) dissolved in water lowers the freezing point and raises the boiling point of the coolant, preventing freezing in winter and overheating in summer.
  • Ice cream making: Salt (or CaCl2) added to ice around the ice-cream can lowers the melting/freezing point of the ice-salt mixture, allowing the mixture to reach temperatures below 0 °C to freeze the ice-cream mixture.
  • Cooking: Adding salt slightly elevates the boiling point of water, but typical culinary amounts change the boiling point negligibly (not the main reason salted water cooks faster).
  • Seawater: Ocean water freezes at about -1.8 °C instead of 0 °C because of dissolved salts (freezing point depression).
🧮 Formulas
  1. \[Raoult's law (for solvent in ideal solution): p_solvent = x_solvent · p°_solvent\]
  2. \[Vapor-pressure lowering (dilute): Δp = p°_solvent - p_solvent ≈ x_solute · p°_solvent\]
  3. \[Boiling point elevation (non-electrolyte): ΔTb = Kb · m\]
  4. \[Freezing point depression (non-electrolyte): ΔTf = Kf · m\]
  5. \[With electrolytes (van't Hoff factor i): ΔT = i · K · m (so ΔTb = i · Kb · m, ΔTf = i · Kf · m)\]
  6. \[Solution boiling/freezing temperature: Tb(solution) = Tb(solvent) + ΔTb\]
    \[Tf(solution) = Tf(solvent) - ΔTf\]
🎈6

Osmotic Pressure

Fig 6 — Educational Diagram: Osmotic Pressure

Fig 6 — Educational Diagram: Osmotic Pressure

⚗️ CHEMICAL REACTION

Osmotic Pressure

Core Principle: πV = n_solute RT (for ideal dilute solutions)

Definition: Osmotic pressure (π) is the external pressure that must be applied to a solution to stop the net flow of solvent through a semipermeable membrane from the pure solvent into the solution. It is a colligative property — it depends on the number of solute particles, not their identity.

Qualitative explanation: If a semipermeable membrane separates pure solvent (e.g., water) and a solution, solvent molecules tend to move into the solution side (where solvent chemical potential is lower) until equilibrium is reached. To stop this flow, an external pressure π is applied on the solution side. That pressure is the osmotic pressure.

Thermodynamic derivation (concise): At equilibrium, the chemical potential of the solvent in the pure state (μ1° at pressure P) equals the chemical potential of the solvent in the solution under the applied pressure (μ1 at pressure P+π). For a dilute ideal solution:

  • μ1(solution, P+π) = μ1°(P) + V̄1 π + RT ln x1
  • Setting μ1°(P) = μ1(solution, P+π) and using x1 ≈ 1 − n2/n1 (dilute), we get: V̄1 π ≈ −RT ln x1 ≈ RT (n2/n1).

For 1 L of solution where n1 ≈ (mass of solvent)/M1 and V ≈ volume of solution, this simplifies to the van't Hoff form:

πV = n2RT or equivalently π = CRT where C is molar concentration of solute (mol·L−1) and R is the gas constant. For electrolytes, use the van't Hoff factor i: π = iMRT.

Assumptions & limits: The van't Hoff relation is valid for dilute, ideally behaving solutions. Deviations occur at higher concentrations due to solute-solute and solute-solvent interactions; electrolytes may not fully dissociate (effective i < expected) and activity coefficients become important.

How it’s used: Osmotic pressure is used to determine molar masses of large molecules (proteins, polymers) by measuring π at known concentration (osmometry). It is also important in biology (cell volume regulation), water purification (reverse osmosis), and medicine (infusion fluids, dialysis).

📌 Examples
  • Biology: Red blood cells placed in pure water swell and can burst (hemolysis) because water moves into the cell by osmosis; in a hypertonic saline solution they shrink (crenation).
  • Plants: Plasmolysis occurs when a plant cell in a hypertonic solution loses water and the cell membrane pulls away from the cell wall.
  • Desalination: Reverse osmosis applies pressure greater than the osmotic pressure to force water from saline solution through a semipermeable membrane to obtain freshwater.
  • Medical dialysis: Artificial kidneys remove waste by allowing diffusion and controlling osmotic gradients across membranes.
  • Analytical chemistry: Determination of molar mass of high-molecular-weight solutes (proteins, polymers) by measuring osmotic pressure at known concentration.
🧮 Formulas
  1. \[πV = n_solute RT (for ideal dilute solutions)\]
  2. \[π = CRT (C = molar concentration\]
    \[R = 0.082057 L·atm·K−1·mol−1\]
    \[T in K)\]
  3. \[π = iMRT (i = van't Hoff factor for electrolytes)\]
  4. \[Molar mass from osmotic pressure: M = (wRT)/(πV) where w = mass of solute\]
    \[V = solution volume\]
  5. \[Approximate dilute relation: π ≈ (n2 / n1) (RT / V̄1) leading to π ∝ concentration\]
🔬7

Determination of Molar Mass from Colligative Properties

Fig 7 — Educational Diagram: Determination of Molar Mass from Colligative Properties

Fig 7 — Educational Diagram: Determination of Molar Mass from Colligative Properties

⚗️ CHEMICAL REACTION

Determination of Molar Mass from Colligative Properties

Core Principle: ΔT_f = K_f · m (freezing point depression)

What are colligative properties? Colligative properties depend only on the number of solute particles (not their nature). Important colligative properties are lowering of vapor pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. These properties can be used experimentally to determine the molar mass (molecular weight) of a solute.

Basic idea: For a known mass of solute dissolved in a known mass (or volume) of solvent, measure a colligative effect (e.g., ΔT_f, ΔT_b, ΔP, or Π). Use the relevant relation to find the number of moles of solute; then molar mass = (mass of solute)/(moles of solute).

Common methods and their relations (for dilute, ideally behaving solutions; non-volatile solute):

  • Freezing point depression: ΔT_f = K_f · m (m = molality = mol solute/kg solvent). Measure ΔT_f, compute molality m = ΔT_f / K_f, then moles solute = m · (kg solvent), and molar mass = mass_solute / moles_solute.
  • Boiling point elevation: ΔT_b = K_b · m (analogue of freezing case).
  • Osmotic pressure: Π = MRT (M = molarity in mol L⁻¹, R = gas constant, T = K). Measure Π, find M = Π/(RT), moles = M · volume (L), then molar mass = mass_solute / moles.
  • Vapour pressure lowering (Raoult's law): ΔP = X_solute · P°_solvent and ΔP/P° = X_solute ≈ n_solute / n_solvent for dilute solutions. Using masses and known solvent molar mass one can solve for solute molar mass.

Corrections and assumptions:

  • For electrolytes that dissociate or for solutes that associate, include the van't Hoff factor i: effective molality = i · m and ΔT = i K m, Π = i MRT, etc.
  • Assume dilute solutions and non-volatile solute (so solvent vapour pressure measured is that of the solvent only) and ideal behaviour.

Typical experimental procedure (freezing/boiling point method):

  1. Measure the freezing (or boiling) point of pure solvent.
  2. Dissolve a known mass of solute in a known mass of solvent; measure the new freezing (or boiling) point.
  3. Compute ΔT = T_pure - T_solution (for freezing) or ΔT = T_solution - T_pure (for boiling).
  4. Find molality m = ΔT / K_f (or ΔT / K_b). Compute moles solute = m · kg_solvent. Finally, molar mass = mass_solute / moles_solute.

Why it works (short derivation highlights): Freezing/boiling point shifts arise because solute particles change the chemical potential of the solvent; for dilute solutions these shifts are proportional to molality with proportionality constants (K_f, K_b) that depend on the solvent. Osmotic pressure comes from solvent flow tendency and equals the pressure required to stop solvent flow; for dilute ideal solutions Π = nRT/V = MRT.

📌 Examples
  • Example 1 — Freezing point depression: 1.25 g of an unknown solid is dissolved in 25.00 g benzene. The freezing point of the solution is 1.92 K lower than pure benzene. Given K_f(benzene) = 5.12 K·kg·mol⁻¹. Find molar mass of the unknown. Solution: m = ΔT/K_f = 1.92/5.12 = 0.375 mol·kg⁻¹. Moles solute = m · mass_solvent(kg) = 0.375 × 0.02500 = 0.009375 mol. Molar mass = mass/moles = 1.25 g / 0.009375 mol = 133.3 g·mol⁻¹.
  • Example 2 — Osmotic pressure: 0.500 g of a protein is dissolved in 200.0 mL water at 25 °C. The measured osmotic pressure is 0.0248 atm. Find the molar mass. Solution: Molarity M = Π/(RT) = 0.0248 / (0.08206 × 298) = 0.001015 mol·L⁻¹. Moles in 0.200 L = 0.001015 × 0.200 = 0.0002030 mol. Molar mass = mass / moles = 0.500 g / 0.0002030 mol ≈ 2463 g·mol⁻¹.
  • Practical example — Antifreeze: Ethylene glycol added to automobile coolant lowers the freezing point (and raises boiling point) of water. By measuring the freezing point depression for a known composition, the concentration (and thus effective molar fraction) of ethylene glycol can be inferred.
  • Practical example — Determining molar mass of macromolecules: Osmotic pressure or freezing point depression are commonly used to estimate molar masses of polymers and proteins because these colligative methods respond to particle number, allowing calculation of very large molar masses.
🧮 Formulas
  1. \[ΔT_f = K_f · m (freezing point depression)\]
  2. \[ΔT_b = K_b · m (boiling point elevation)\]
  3. \[m = molality = moles solute / kg solvent\]
  4. \[Π = M R T (osmotic pressure\]
    \[M = molarity in mol·L⁻¹\]
    \[R = 0.08206 L·atm·K⁻¹·mol⁻¹)\]
  5. \[ΔP = X_solute · P°_solvent (vapour pressure lowering\]
    \[ΔP/P° = X_solute ≈ n_solute/n_solvent for dilute solutions)\]
  6. \[Include van't Hoff factor for dissociation/association: ΔT = i K m, Π = i M R T\]
🔬8

Van't Hoff Factor and Abnormal Molar Masses

Fig 8 — Educational Diagram: Van

Fig 8 — Educational Diagram: Van't Hoff Factor and Abnormal Molar Masses

⚗️ CHEMICAL REACTION

Van't Hoff Factor and Abnormal Molar Masses

Core Principle: Van't Hoff factor (dissociation): i = 1 + α (v − 1)

Van't Hoff factor (i) — definition: Van't Hoff factor i is the ratio of the actual number of solute particles in solution (taking dissociation or association into account) to the number of formula units dissolved. It is used to correct colligative property equations for electrolytes or associating solutes.

Why it matters: Colligative properties (boiling point elevation, freezing point depression, osmotic pressure, vapor-pressure lowering) depend on the number of solute particles, not their identity. If a solute dissociates into ions (electrolyte) or associates into larger units (dimerisation), the effective number of particles changes and the simple non-electrolyte formulas must be multiplied by i.

Derivation and common cases:

- Dissociation: if one formula unit dissociates into v particles (for example NaCl -> Na+ + Cl- has v = 2), and fraction α dissociates, the total number of particles per initial formula unit is (1 − α) + α·v = 1 + α(v − 1). So

i = 1 + α (v − 1)

From this, α = (i − 1)/(v − 1).

- Association (k-mers): if k monomers associate to form one particle and fraction α of monomers associate, the particle count per initial monomer becomes (1 − α) + α/k = 1 − α(1 − 1/k). So

i = 1 − (k − 1)/k · α = 1 − α(1 − 1/k)

For dimerisation (k = 2) this gives i = 1 − α/2.

Effect on colligative properties (dilute solution approximations):

  • Freezing point depression: ΔTf = i · Kf · m
  • Boiling point elevation: ΔTb = i · Kb · m
  • Osmotic pressure: Π = i · M · R · T
  • Relative lowering of vapour pressure (for dilute solutions): ΔP/P ≈ i · x2 (x2 = mole fraction of solute)

Abnormal molar masses (apparent molar mass from colligative measurements):

When we determine molar mass from a colligative property assuming the solute gives 1 particle per formula unit (i = 1), but the solute actually dissociates or associates, the experimentally determined molar mass M_exp will differ from the true molar mass M_true. From Π = i M_true R T (n/V) and the naive formula M_exp = (mass · R · T)/(Π · V) (which assumes i = 1) we get

M_exp = M_true / i

Thus:

  • If dissociation occurs (i > 1): M_exp < M_true (apparent molar mass is smaller).
  • If association occurs (i < 1): M_exp > M_true (apparent molar mass is larger).

Estimating degree of dissociation/association from experiments:

- If M_true (from molecular formula) and M_exp (from a colligative measurement) are known, then i = M_true / M_exp. Use i in the dissociation/association formulas above to find α.

Limitations and real behaviour:

- The ideal values of i (equal to v for complete dissociation or 1/k for complete association) are approached only at infinite dilution. In real solutions inter-ionic attractions, ion-pairing and incomplete dissociation make observed i lower than the ideal value for electrolytes. i depends on concentration, solvent, temperature and ionic strength.

Practical importance / real-life examples: salts in seawater change boiling/freezing points and osmotic pressure; association of weak acids in non-polar solvents (e.g., acetic acid in benzene) affects colligative-based molar-mass determinations; understanding i is essential in biochemical and industrial processes involving osmotic balance.

Summary: always correct colligative equations by the van't Hoff factor i. Use i = 1 + α(v − 1) for dissociation and i = 1 − α(1 − 1/k) for association. Apparent (abnormal) molar mass relates to true molar mass by M_exp = M_true / i.

📌 Examples
  • Example 1 (dissociation): NaCl → Na+ + Cl- has v = 2. If degree of dissociation α = 0.9, i = 1 + 0.9(2 − 1) = 1.9. Freezing point depression = i · Kf · m = 1.9 · Kf · m.
  • Example 2 (association): Acetic acid in benzene dimerises (k = 2). If 80% of monomers are dimerised (α = 0.8), i = 1 − α/2 = 1 − 0.8/2 = 0.6. A molar mass measured from vapor-pressure lowering would appear M_exp = M_true / 0.6 ≈ 1.67 · M_true (i.e. abnormally high).
  • Example 3 (find α from experimental molar mass): A solute with formula mass 100 g·mol−1 gives an apparent molar mass 50 g·mol−1 from osmotic pressure. Then i = 100/50 = 2. If it dissociates into v = 3 ions, α = (i − 1)/(v − 1) = (2 − 1)/(3 − 1) = 1/2 (50% dissociation).
  • Example 4 (real behaviour): Ideal complete dissociation of CaCl2 (v = 3) would give i = 3 at infinite dilution, but at practical concentrations measured i might be ≈ 2.3–2.6 due to ion pairing and inter-ionic attraction.
🧮 Formulas
  1. \[Van't Hoff factor (dissociation): i = 1 + α (v − 1)\]
  2. \[Van't Hoff factor (association\]
    \[k-mers): i = 1 − α (1 − 1/k)\]
  3. \[Degree of dissociation from i: α = (i − 1)/(v − 1)\]
  4. \[Degree of association from i (k-mers): α = (1 − i) · k/(k − 1)\]
  5. \[Freezing point depression: ΔTf = i · Kf · m\]
  6. \[Boiling point elevation: ΔTb = i · Kb · m\]
🧴9

Non-ideal Solutions and Deviations from Raoult's Law

Fig 9 — Educational Diagram: Non-ideal Solutions and Deviations from Raoult

Fig 9 — Educational Diagram: Non-ideal Solutions and Deviations from Raoult's Law

📜 THEOREM / LAW

Non-ideal Solutions and Deviations from Raoult's Law

Core Principle: Raoult's law (ideal): p_A = x_A p_A^* , p_B = x_B p_B^*

What are non‑ideal solutions? A non‑ideal solution is one in which the interactions between unlike molecules (A–B) differ significantly from those between like molecules (A–A or B–B). Because of this, the vapour pressures and other thermodynamic properties do not follow Raoult's law exactly.

Raoult's law (ideal behaviour): For an ideal binary liquid solution of components A and B, the partial vapour pressure of each component is proportional to its mole fraction in the liquid:

p_A = x_A p_A^* ,   p_B = x_B p_B^*

and total vapour pressure

p_total = x_A p_A^* + x_B p_B^*

Here p_A^* and p_B^* are the vapour pressures of the pure components at the temperature of interest.

Why deviations occur:

  • If A–B interactions are weaker than A–A and B–B, molecules escape more easily → vapour pressure higher than predicted (positive deviation).
  • If A–B interactions are stronger than A–A and B–B, molecules are held more tightly → vapour pressure lower than predicted (negative deviation).

Character of deviations and consequences:

  • Positive deviation: total vapour pressure is greater than the Raoult's‑law value; boiling point is lower than expected. Can produce a minimum‑boiling azeotrope (constant‑boiling mixture) which cannot be separated by simple distillation beyond the azeotropic composition.
  • Negative deviation: total vapour pressure is less than Raoult's‑law value; boiling point is higher than expected. Can produce a maximum‑boiling azeotrope.

Activity and activity coefficient (measure of non‑ideality):

Introduce activity a_A so that p_A = a_A p_A^*. For liquid solutions a_A is often written as a_A = x_A γ_A, where γ_A is the activity coefficient. For an ideal solution γ_A = 1. Deviations correspond to γ_A ≠ 1 (γ_A > 1 for positive deviation, γ_A < 1 for negative deviation).

Vapour composition: The mole fraction of A in vapour (y_A) is

y_A = p_A / p_total = x_A γ_A p_A^* / (x_A γ_A p_A^* + x_B γ_B p_B^*)

Azeotrope condition (brief): An azeotrope is a composition where liquid and vapour compositions are identical (y_A = x_A). In practice, azeotropes appear where the p_total vs composition curve has an extremum (dp_total/dx = 0), giving constant‑boiling behaviour.

Significance / real life:

  • Industrial distillation: azeotropes (e.g., ethanol–water) limit the purity obtainable by simple distillation and require special separation methods (azeotropic distillation, extractive distillation, molecular sieves).
  • Formulation of mixtures (solvents, fuels): non‑ideal behaviour affects vapour release, flammability, and evaporation rates.

How to detect non‑ideality experimentally: Plot partial or total vapour pressure versus composition and compare with straight lines from Raoult's law; deviations above the line indicate positive deviation, below indicate negative deviation. Measure activity coefficients from p_A/(x_A p_A^*).

Summary: Non‑ideal solutions occur when intermolecular forces change on mixing. Raoult's law is recovered when these forces are similar. Deviations are quantified by activity coefficients and give rise to practical phenomena such as azeotropes which strongly influence separation processes.

📌 Examples
  • Ethanol–water: forms an azeotrope (≈95.6% ethanol) that limits the purity obtainable by simple distillation — important in producing absolute ethanol.
  • Chloroform–acetone: strong specific interactions (H‑bonding / donor–acceptor) cause negative deviation from Raoult's law and lead to non‑ideal vapour behaviour.
  • Benzene–toluene or hexane–heptane: chemically similar pairs that behave nearly ideally and closely follow Raoult's law (used as examples of ideal‑like behaviour).
  • Ethanol–cyclohexane: dissimilar polarities give weaker A–B interactions and typically show positive deviation (vapour pressure higher than ideal).
🧮 Formulas
  1. \[Raoult's law (ideal): p_A = x_A p_A^*\]
    \[p_B = x_B p_B^*\]
  2. \[Total pressure: p_total = x_A p_A^* + x_B p_B^*\]
  3. \[Activity form: p_A = a_A p_A^* = x_A γ_A p_A^* (γ_A = activity coefficient)\]
  4. \[Vapour composition: y_A = p_A / p_total = x_A γ_A p_A^* / (x_A γ_A p_A^* + x_B γ_B p_B^*)\]
  5. \[Azeotrope condition (practical): liquid and vapour compositions equal → y_A = x_A\]
    \[equivalently\]
    \[extremum in p_total vs x (dp_total/dx = 0)\]
  6. \[Deviation indicator: γ_i = p_i / (x_i p_i^*): γ_i &gt\]
    \[1 (positive deviation), γ_i &lt\]
    \[1 (negative deviation)\]
💨10

Solubility of Gases in Liquids and Henry's Law

Fig 10 — Educational Diagram: Solubility of Gases in Liquids and Henry

Fig 10 — Educational Diagram: Solubility of Gases in Liquids and Henry's Law

📜 THEOREM / LAW

Solubility of Gases in Liquids and Henry's Law

Core Principle: Henry's law (mole-fraction form): p = k_H · x (p = partial pressure; x = mole fraction of gas in liquid; k_H has units of pressure)

Definition (Henry's law - CBSE statement): At a constant temperature, the solubility (expressed as mole fraction) of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically: p = kH x, where p is the partial pressure of the gas, x is its mole fraction in the liquid and kH is Henry's law constant (for that gas–solvent pair at that temperature).

Alternate concentration form: When concentration is used (molarity c, mol·L−1), Henry's law is commonly written as c = k'H p, where k'H (often called the Henry's law solubility coefficient) = c/p and has units such as mol·L−1·atm−1. Different authors swap which constant is named kH, so check the definition used.

Why it holds (brief physical picture): At equilibrium the tendency of gas molecules to escape from the liquid equals the tendency of gas molecules in the gas phase to dissolve. For relatively low concentrations and non-reacting gases, this balance leads to a linear relation between dissolved amount and partial pressure.

Applicability and limitations: Henry's law is valid for dilute solutions of non-reacting gases and at moderate pressures. Deviations occur at high pressures, high concentrations, or when the gas chemically reacts with the solvent (e.g., CO2 reacts with water to form H2CO3), which increases apparent solubility beyond Henry's law prediction.

Factors affecting gas solubility:

  • Pressure: Solubility ∝ partial pressure of the gas (Henry's law). Increasing pressure increases solubility linearly (for ideal conditions).
  • Temperature: For most gases dissolving in liquids the process is exothermic, so solubility decreases as temperature increases. This is why warm soda goes flat faster.
  • Nature of gas and solvent: Nonpolar gases dissolve better in nonpolar solvents; polar gases (or those that react) behave differently. CO2 is more soluble in water than O2 because it reacts partially with water.
  • Presence of dissolved salts (salting-out): Addition of salts usually decreases gas solubility in water ("salting-out").

Temperature dependence (van 't Hoff type relation): The Henry's law constant is temperature dependent. Using equilibrium thermodynamics one can write a van 't Hoff form:
ln(kH(T2)/kH(T1)) = -ΔHsol/R · (1/T2 - 1/T1)
or equivalently a differential form d ln kH/dT = ΔHsol/(R T2), where ΔHsol is the enthalpy change on dissolution. For exothermic dissolution (ΔHsol < 0) kH decreases with increasing T, i.e. solubility falls with rising temperature.

Key conceptual points to remember:

  • Henry's law links gas partial pressure and amount dissolved (linear relation at low concentrations).
  • Different conventions exist for kH — check whether p = kH x or c = k'H p is used.
  • Reactive gases and high-pressure conditions cause deviations. Chemical reactions (e.g., CO2 + H2O ⇄ H2CO3) increase apparent solubility.

📌 Examples
  • Carbonated drinks (soda/beer): CO2 is dissolved in the liquid under high pressure. When the bottle is opened pressure above the liquid falls, CO2 solubility drops and bubbles form (escape of gas).
  • Decompression sickness (the 'bends'): Rapid decrease in ambient pressure (as in a diver surfacing too quickly) causes N2 dissolved in blood/tissues to come out of solution as bubbles, causing pain and danger.
  • Dissolved oxygen in water: Oxygen solubility in water decreases with increasing temperature; cold water holds more O2 and supports aquatic life better.
  • CO2 uptake by oceans: Increased atmospheric CO2 increases the partial pressure of CO2 above seawater so more CO2 dissolves (and reacts), contributing to ocean acidification.
🧮 Formulas
  1. \[Henry's law (mole-fraction form): p = k_H · x (p = partial pressure\]
    \[x = mole fraction of gas in liquid\]
    \[k_H has units of pressure)\]
  2. \[Henry's law (concentration form): c = k'_H · p (c = concentration in mol·L^-1\]
    \[p in atm\]
    \[k'_H in mol·L^-1·atm^-1)\]
  3. \[Conversion note: k_H (pressure units) = p/x\]
    \[k'_H (solubility coefficient) = c/p\]
    \[Check which convention is used.\]
  4. \[Temperature dependence (van 't Hoff form): ln(k_H(T2)/k_H(T1)) = -ΔH_sol/R · (1/T2 - 1/T1)\]
    \[where ΔH_sol is enthalpy of dissolution and R is gas constant\]
  5. \[For small concentration/low pressure the relation is linear: plot of dissolved amount vs partial pressure is a straight line through origin.\]
🔬11

Applications, Problem-Solving Strategies and Experimental Methods

Fig 11 — Educational Diagram: Applications, Problem-Solving Strategies and Experimental Methods

Fig 11 — Educational Diagram: Applications, Problem-Solving Strategies and Experimental Methods

⚗️ CHEMICAL REACTION

Applications, Problem-Solving Strategies and Experimental Methods

Core Principle: Raoult's law: P_solution = x_solvent · P°_solvent

Overview
This topic deals with colligative properties of dilute solutions (vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure), their applications, typical problem-solving approaches, and experimental methods used to measure these properties and determine molar masses.

Core concepts

  • Colligative properties depend on the number of solute particles, not their identity (for ideal dilute solutions).
  • Raoult's law (for ideal solutions): P_solution = x_solvent · P°_solvent. For a non-volatile solute, vapor pressure lowering = P° − P = x_solute · P°.
  • Boiling point elevation and freezing point depression are proportional to solute molality (m): ΔT_b = K_b · m and ΔT_f = K_f · m.
  • Osmotic pressure π for dilute solutions: π = C · R · T (C = molarity). Equivalent: πV = nRT.
  • Electrolytes dissociate; introduce Van't Hoff factor i (effective number of particles). For dissociation, ΔT = iK·m. For partial dissociation, i = 1 + (z − 1)α for a salt that gives z ions and dissociation fraction α.

Problem-solving strategies

  • Read carefully and list given data (mass, temperature changes, densities, volumes, molar masses if known).
  • Decide which colligative property applies (ΔT_b, ΔT_f, vapor pressure lowering, or π) and pick the appropriate formula.
  • Use molality (m = moles solute / kg solvent) for ΔT_b and ΔT_f; use molarity (C) for osmotic pressure. Convert volumes → mass → moles as needed.
  • For electrolyte solutions include Van't Hoff factor i. If ionization/dissociation given indirectly, relate i and degree of dissociation α.
  • For vapor pressure problems use mole fractions: x_solute = n_solute / (n_solute + n_solvent). For dilute solutions x_solute ≈ n_solute/(n_solvent) when n_solute << n_solvent (use with caution).
  • Check units (Kb, Kf are in K·kg·mol−1, R usually 0.08206 L·atm·K−1·mol−1 or 8.314 J·mol−1·K−1) and convert temperatures to K where needed (for π = CRT temperature must be in K).
  • Use approximations only when justified (very dilute solutions behave ideally). For concentrated or strongly interacting systems expect deviations from Raoult's law.

Experimental methods & practical procedure

  • Ebullioscopy (boiling point elevation): Measure increase in boiling point of solvent after adding known mass of solute. Use ΔT_b = K_b · m to compute molar mass of solute. Apparatus: ebulliometer (boiling point thermometer, sample bulb, heating and reflux to ensure equilibrium).
  • Cryoscopy (freezing point depression): Determine depression of freezing point ΔT_f to calculate molar mass using ΔT_f = K_f · m. Apparatus: cryoscope, controlled cooling bath, thermometer.
  • Osmometry (osmotic pressure): Place solution and pure solvent separated by semipermeable membrane; measure pressure required to stop osmosis. Use π = CRT to find molar mass (from π and concentration). Practical for macromolecules (proteins, polymers).
  • Vapour-pressure lowering and isopiestic methods: Measure vapor pressure of solution relative to pure solvent or equilibrate with a reference solution of known properties. Apply Raoult's law to determine mole fractions and molar mass.
  • Typical workflow when determining molar mass experimentally: weigh solute, dissolve in measured solvent mass/volume, measure colligative effect, compute m (or C), then deduce molar mass from n = mass / M and relation between n and measured effect.
  • Common sources of error: non-ideality (assume dilute), incomplete dissociation, temperature measurement inaccuracies, volatile solutes (affect vapor pressure), impurities in solvent.

Applications — real life

  • Antifreeze in automobile radiators (ethylene glycol lowers freezing point and raises boiling point of coolant).
  • Desalination and water purification via reverse osmosis (osmotic pressure concept).
  • Blood osmolarity and medical IV solutions (isotonic saline to avoid cell swelling or shrinkage).
  • Determination of molar masses of polymers and biomolecules via osmometry or boiling/freezing point methods.
  • Carbonated beverages: CO2 solubility follows Henry's law; pressure and temperature control carbonation.

Tips & quick checks

  • Molality is temperature-independent (useful for boiling/freezing experiments); molarity is temperature-dependent.
  • Use molality for ΔT calculations and molarity for osmotic pressure problems (or convert appropriately).
  • At very low concentrations Raoult's and Henry's laws are good approximations; at higher concentrations expect positive/negative deviations.
📌 Examples
  • Antifreeze: Adding ethylene glycol to water lowers freezing point — application of ΔT_f = K_f · m; typical calculation finds required mass of ethylene glycol to lower freezing point to a target temperature.
  • Determining molar mass by freezing point depression: Dissolve known mass of unknown solute in 100 g water, measure ΔT_f = 1.86 K, use K_f(water)=1.86 K·kg·mol−1 => m = ΔT_f/K_f = 1.00 mol·kg−1 → moles solute = 0.100 mol → M = mass/moles.
  • Osmotic pressure for macromolecules: A polymer solution has π = 0.050 atm at 25 °C and concentration 0.00100 mol·L−1 → M can be found by π = CRT with R = 0.08206 L·atm·K−1·mol−1 and T in K.
  • Vapour pressure lowering: For a solution made by dissolving a nonvolatile solute, measured vapor pressure P is 95.0 kPa while P° of pure solvent is 100.0 kPa → relative lowering = (P°−P)/P° = 0.05 = x_solute, use mole count to get molar mass of solute.
  • Electrolyte effect: 0.10 mol·kg−1 NaCl produces larger ΔT_f than 0.10 mol·kg−1 glucose because NaCl dissociates into 2 ions (i ≈ 2), so effective molality doubles (ΔT ∝ i·m).
  • Henry's law and carbonation: Solubility of CO2 in water increases with pressure; s = k_H·P(CO2) — relevant to how soda retains fizz under sealed pressure.
🧮 Formulas
  1. \[Raoult's law: P_solution = x_solvent · P°_solvent\]
  2. \[Relative lowering of vapour pressure: (P° − P)/P° = x_solute (for nonvolatile solute)\]
  3. \[Boiling point elevation: ΔT_b = K_b · m\]
  4. \[Freezing point depression: ΔT_f = K_f · m\]
  5. \[Osmotic pressure: π = C · R · T (or πV = nRT)\]
  6. \[Molality: m = moles of solute / mass of solvent (kg)\]

Key Concepts

Solution
A homogeneous mixture of two or more substances where the solute is uniformly dispersed in the solvent.
Solute
The substance that is dissolved in a solvent to form a solution.
Solvent
The component of a solution present in the largest amount; it dissolves the solute.
Solubility
The maximum amount of a solute that can dissolve in a specified amount of solvent at a given temperature, usually expressed per 100 g of solvent.
Saturated / Unsaturated / Supersaturated (solution states)
Unsaturated: contains less solute than can dissolve; Saturated: contains maximum solute at equilibrium; Supersaturated: contains more solute than equilibrium (metastable).
Molarity (M)
Concentration expressed as moles of solute per litre of solution (mol L⁻¹).
Molality (m)
Concentration expressed as moles of solute per kilogram of solvent (mol kg⁻¹).
Mole fraction (χ)
Ratio of moles of a component to the total moles of all components in the mixture.
Mass percent (w/w %)
Mass of solute divided by mass of solution, multiplied by 100.
Parts per million (ppm)
A concentration unit equal to one part solute per 10^6 parts solution (mg per kg for dilute aqueous systems).
Raoult's law
For an ideal solution, the partial vapour pressure of each volatile component equals the vapour pressure of the pure component multiplied by its mole fraction in the solution: P_i = x_i P_i°.
Henry's law
At a fixed temperature, the solubility (or concentration) of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid: c = k_H · p.
Ideal solution
A solution that obeys Raoult's law at all compositions; interactions between unlike molecules equal those between like molecules.
Colligative properties
Properties of solutions that depend only on the number of solute particles (not their identity): vapour pressure lowering, boiling point elevation, freezing point depression, osmotic pressure.
Vapour pressure lowering
Decrease in the solvent's vapour pressure on adding a non-volatile solute; for ideal dilute solutions ΔP = x_solute · P°_solvent.
Boiling point elevation
Increase in the boiling temperature of a solvent on adding a solute; quantitatively ΔT_b = K_b · m · i (molal boiling point constant K_b, molality m, van't Hoff factor i).
Freezing point depression
Lowering of the freezing point of a solvent by the addition of a solute; ΔT_f = K_f · m · i (K_f is molal freezing point constant).
Osmotic pressure (Π)
Pressure required to stop solvent flow through a semipermeable membrane into a solution; for ideal dilute solutions Π = i·M·R·T.
van't Hoff factor (i)
The ratio of the number of particles in solution after dissociation or association to the number of formula units dissolved; accounts for electrolytic dissociation in colligative properties.
Normality (N)
Concentration expressed as gram-equivalents of solute per litre of solution; N = equivalents per litre. Equivalent depends on the reaction (acid–base, redox).

Practice Questions

  1. Define molality and explain why it is preferred over molarity for studying colligative properties. / मोललता को परिभाषित कीजिए और समझाइए कि अणुसंख्यक गुणों के अध्ययन हेतु इसे मोलरता पर वरीयता क्यों दी जाती है।
    Show answer

    Molality is the moles of solute per kilogram of solvent; it is preferred because, unlike molarity, it does not depend on volume and hence is independent of temperature. / मोललता प्रति किलोग्राम विलायक में विलेय के मोलों की संख्या है; इसे वरीयता दी जाती है क्योंकि मोलरता के विपरीत यह आयतन पर निर्भर नहीं करती और इसलिए तापमान से स्वतंत्र होती है।

  2. State Raoult's law for a solution containing a non-volatile solute and write the expression for relative lowering of vapour pressure. / अवाष्पशील विलेय युक्त विलयन के लिए राउल्ट का नियम बताइए और वाष्प दाब के आपेक्षिक अवनमन का व्यंजक लिखिए।
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    For a non-volatile solute, p = x_solvent · p°_solvent, and the relative lowering of vapour pressure Δp/p° = x_solute, which is a colligative property. / अवाष्पशील विलेय हेतु p = x_solvent · p°_solvent, और वाष्प दाब का आपेक्षिक अवनमन Δp/p° = x_solute होता है, जो एक अणुसंख्यक गुण है।

  3. Why does the solubility of a gas in a liquid decrease with rise in temperature? / द्रव में गैस की विलेयता तापमान बढ़ने पर क्यों घटती है?
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    Dissolution of a gas is exothermic, so increasing temperature shifts the dissolving equilibrium backward, driving gas out of solution and lowering its solubility. / गैस का घुलना ऊष्माक्षेपी होता है, अतः तापमान बढ़ने पर घुलने का साम्य पीछे की ओर खिसकता है, जिससे गैस विलयन से बाहर निकलती है और उसकी विलेयता घटती है।

  4. 1.25 g of an unknown solid in 25.00 g benzene lowers the freezing point by 1.92 K (K_f = 5.12 K kg mol⁻¹). Find its molar mass. / 25.00 g बेंजीन में 1.25 g अज्ञात ठोस हिमांक को 1.92 K घटाता है (K_f = 5.12 K kg mol⁻¹)। इसका मोलर द्रव्यमान ज्ञात कीजिए।
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    m = ΔT_f/K_f = 1.92/5.12 = 0.375 mol kg⁻¹; moles = 0.375 × 0.025 = 0.009375; molar mass = 1.25/0.009375 ≈ 133.3 g mol⁻¹. / m = ΔT_f/K_f = 1.92/5.12 = 0.375 mol kg⁻¹; मोल = 0.375 × 0.025 = 0.009375; मोलर द्रव्यमान = 1.25/0.009375 ≈ 133.3 g mol⁻¹।

  5. Distinguish between positive and negative deviations from Raoult's law in terms of intermolecular interactions. / अंतराआण्विक अंतःक्रियाओं के संदर्भ में राउल्ट के नियम से धनात्मक और ऋणात्मक विचलनों में अंतर कीजिए।
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    Positive deviation occurs when A–B interactions are weaker than A–A and B–B, raising vapour pressure above ideal, while negative deviation occurs when A–B interactions are stronger, lowering vapour pressure below ideal. / धनात्मक विचलन तब होता है जब A–B अंतःक्रियाएँ A–A और B–B से दुर्बल हों, जिससे वाष्प दाब आदर्श से अधिक हो जाता है, जबकि ऋणात्मक विचलन तब होता है जब A–B अंतःक्रियाएँ प्रबल हों, जिससे वाष्प दाब आदर्श से कम हो जाता है।

  6. Write the formula for osmotic pressure and explain its use in determining molar mass of polymers. / परासरण दाब का सूत्र लिखिए और बहुलकों का मोलर द्रव्यमान ज्ञात करने में इसके उपयोग को समझाइए।
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    Osmotic pressure π = MRT (or iMRT); molar mass M = wRT/(πV), and because osmotic pressure is large and measurable even at low concentration, it is ideal for finding the high molar masses of polymers. / परासरण दाब π = MRT (या iMRT); मोलर द्रव्यमान M = wRT/(πV), और चूँकि परासरण दाब कम सांद्रता पर भी बड़ा और मापने योग्य होता है, यह बहुलकों के उच्च मोलर द्रव्यमान ज्ञात करने हेतु आदर्श है।

  7. Define the van't Hoff factor and calculate its value for NaCl that is 90% dissociated. / वान्ट हॉफ कारक को परिभाषित कीजिए और 90% वियोजित NaCl के लिए इसका मान ज्ञात कीजिए।
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    The van't Hoff factor i is the ratio of actual particles to formula units dissolved; for NaCl, i = 1 + α(v − 1) = 1 + 0.9(2 − 1) = 1.9. / वान्ट हॉफ कारक i वास्तविक कणों और घुले सूत्र इकाइयों का अनुपात है; NaCl के लिए i = 1 + α(v − 1) = 1 + 0.9(2 − 1) = 1.9।

  8. Acetic acid dimerises in benzene; explain why this gives an abnormally high observed molar mass. / ऐसीटिक अम्ल बेंजीन में द्विलकीकरण करता है; समझाइए कि इससे प्रेक्षित मोलर द्रव्यमान असामान्य रूप से अधिक क्यों आता है।
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    Association reduces the number of particles so i < 1, and since M_exp = M_true/i, a value of i less than 1 makes the experimentally observed molar mass larger than the true molar mass. / संगुणन कणों की संख्या घटाता है अतः i < 1 होता है, और चूँकि M_exp = M_true/i, i का 1 से कम मान प्रयोगात्मक रूप से प्रेक्षित मोलर द्रव्यमान को वास्तविक मोलर द्रव्यमान से अधिक बना देता है।

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