SF Prep
07 Chemistry Notes 🕒 Updated 2026-10-07
Page 1
Properties and Definition of Solutions & Henry's Law Concepts

Solutions

1. Properties and Definition of Solutions

  • Solution properties reflect properties of each component mixed.
  • Particle size: < 1 nanometer (10−9 m).
  • Composition: One solvent and one or more solutes (e.g., water + sugar + lemon + salt).
  • Solute: The component that dissolves.
  • Solvent: The component that does the dissolving.
  • The physical state (phase) of a solution is determined by the solvent.

2. Identifying Solute and Solvent

  • Criteria 1 (Different Phases): The component which has the same phase as the final solution is the solvent.
  • Criteria 2 (Same Phase): The component present in the greater amount is the solvent.

3. Components of a Solution

  • A solution consists of two primary components: the solute and the solvent.
  • Solvent: The component present in a larger quantity (greater number of moles).
  • Solute: The component dissolved in the solvent, present in a smaller quantity.

Examples of Solute and Solvent Identification :

SolutionSoluteSolvent
TeaSugar, Tea powderWater
ShikanjiLemon juice, SugarWater
Air (O2, N2, CO2)O2, CO2N2
VinegarAcetic acidWater

4. Types of Solution by Physical State

  • Gaseous solutions
  • Liquid solutions
  • Solid solutions

Gaseous Solutions :

SoluteSolventExample
GasGasO2 + N2
LiquidGasChloroform in N2
SolidGasCamphor in N2

Liquid Solutions :

SoluteSolventExample
GasLiquidO2 in water
LiquidLiquidEthanol in water
SolidLiquidGlucose in water

Solid Solutions :

SoluteSolventExample
GasSolidH2 in Pd
LiquidSolidAmalgam of Hg with Na
SolidSolidCu in Au

5. Solubility of Gas in Liquid

  • Solubility: The maximum amount of gas dissolved in a solvent per litre at a particular temperature.

6. Henry's Law Concepts

  • The partial pressure of a gas in the vapour phase (p) is proportional to the mole fraction of the gas in the solution (χ).
  • p = KHχ.
  • Key characteristics:
    • Different gases have different KH values at the same temperature.
    • The value of KH increases with an increase in temperature.
    • KH is a function of the nature of the gas.
    • For a given pressure, a lower value of KH implies higher solubility of the gas in the liquid.

7. Henry's Law & Solubility Trends

  • For gases, more solubility = less KH (Henry's Law constant).
  • KH depends on the nature of the gas. Polar gases (e.g., SO2) have higher solubility and lower KH values compared to non-polar gases (e.g., H2).
  • In a Pgas vs Xgas graph, a steeper slope represents a higher KH value and thus lower solubility.

8. Vapour Pressure of Pure Liquid

  • Molecules in the bulk:
    • Are stable.
    • Have lower energy.
    • Are in equilibrium.
  • Molecules at the surface:
    • Are not in equilibrium.
    • Are less stable.
    • Have high energy.
  • Equilibrium process: Liq ⇌ Vap, where (rate)evap = (rate)cond.
  • Vapour Pressure (V.P.): The pressure exerted by vapours on the surface of the liquid and the walls of the container at a given temperature (T).

9. Effect of Temperature on Vapour Pressure

  • Vapour pressure of a pure liquid depends only on temperature.
  • In general, as Temperature (T) increases, Vapour Pressure (V.P.) increases.
  • The process Liq ⇌ Vap is endothermic (ΔH = positive).
  • For an endothermic reaction, the equilibrium constant (Kp) increases with an increase in temperature. Since Kp = V.P. for this equilibrium, V.P. increases.
log V.P.2V.P.1 = ΔH2.303R [1T1 - 1T2]
  • The relationship between V.P. and T is exponential.
Page 2
Molecular Basis (Maxwell-Boltzmann Distribution) & Raoult's Law

10. Molecular Basis (Maxwell-Boltzmann Distribution)

  • E0 is the minimum energy required for molecules to leave the liquid surface.
  • The Maxwell-Boltzmann distribution shows that increasing the temperature increases the fraction of molecules having energy greater than or equal to E0.
  • Consequently, the rate of evaporation increases, leading to an increase in vapour pressure.
  • Note: While PV = nRT, for vapour pressure P ∝ T is incorrect. The dependence is exponential, not linear.

11. Intermolecular Forces and Vapour Pressure

  • Vapour pressure is inversely proportional to the strength of intermolecular forces.
  • Intermolecular forces ↑ ⇒ V.P. ↓
  • For two liquids A and B at the same temperature, if their intermolecular forces are f1 and f2 respectively:
  • If f1 > f2, then (V.P.)B > (V.P.)A.

12. Strength of Intermolecular Forces

Order of force strength:

Ionic Force > Ion-Dipole > Hydrogen Bond > Dipole-Dipole > Dipole-Induced Dipole > Vander Waal Force
  • Examples:
    • Ionic: Na+:::Cl−
    • Ion-Dipole: Na+---H2O
    • Hydrogen Bond: H2O---H2O---H2O
    • Dipole-Dipole: Acetone (CH3-C(=O)CH3 interaction)
    • Dipole-Induced Dipole: H2O---Br-Br (δ+ δ-)
  • General order: Hydrogen Bonding > Dipole-Dipole > Vander Waal attraction.
  • (Force of attraction)Polar > (Force of attraction)Non-polar.
  • For non-polar molecules: Larger size/surface area (higher molecular weight) leads to stronger forces of attraction.

13. Vapour Pressure of a Solution (Non-volatile Solute)

  • When a non-volatile solute is added to a volatile solvent, the vapour pressure of the solution (Ps) is lower than that of the pure solvent (P°).
  • Reasons for decrease in V.P.:
    • Number of solvent particles per unit surface area is less in the solution.
    • Due to solute-solvent interactions.
P° > Ps

14. Raoult's Law

  • The partial vapour pressure of each component in a solution is proportional to its mole fraction in the solution.
Ps = P° × Xsolvent

15. Vapour Pressure of a Solution of Two Volatile Liquids

  • For a solution of two miscible, volatile liquids A and B:
PA = P°A XA
PB = P°B XB
PT = PA + PB = P°A XA + P°B XB
  • PT = total vapour pressure, P°A/P°B = vapour pressure of pure components, XA/XB = mole fractions in the solution.

16. PT in Terms of Single Mole Fraction

In terms of XA :

PT = P°A XA + P°B (1 − XA)
PT = P°B + XA (P°A − P°B)

In terms of XB :

PT = P°A (1 − XB) + P°B XB
PT = P°A + XB (P°B − P°A)

17. Composition of Vapour Phase

  • For a liquid mixture, the vapour phase composition is determined by the partial pressures of components:
  • Partial pressure of A (PA) = P°A XA
  • Partial pressure of B (PB) = P°B XB
  • Total Vapour Pressure (PT) = PA + PB
YA = PAPT = P°A XAP°A XA + P°B XB
YB = P°B XBP°A XA + P°B XB
  • Ratio of vapour mole fractions:
YAYB = P°A XAP°B XB  ;  YA1 − YA = P°AP°B XA1 − XA

XA = mole fraction of A in liquid phase; YA = mole fraction of A in vapour phase.

18. Vapour Pressure of Binary Solutions

  • For a solution of two volatile components 1 and 2, where x1 and y1 are mole fractions in liquid and vapour phase respectively, the relationship follows a linear plot for 1/x1 vs 1/y1:
Slope = P°1P°2 ; Intercept = P°2 − P°1P°2
Page 3
Distillation of Ideal Solutions & Immiscible Volatile Liquids

19. Distillation of Ideal Solutions

  • Distillation is a process to separate components of a mixture based on differences in volatility.
  • If A is more volatile than B (P°A > P°B), the vapour will be richer in A than the liquid phase.
  • By repeatedly condensing the vapour (which is richer in the more volatile component) and re-boiling, the liquid becomes progressively purer in the more volatile component.

Pressure-Composition Curve :

  • The P-xA graph displays three regions: liquid (L) at the top, vapour (V) at the bottom, and a liquid-vapour equilibrium region (L+V) in between.
  • A zig-zag path on the graph illustrates the distillation process: moving from the liquid composition (xA) to the vapour composition (yA) in the L+V region, which then becomes the liquid composition (x'A) for the next stage.

20. Ideal Solution: Fractional Distillation

  • For the nth container in a fractional distillation process, the mole fraction of component A in the vapour phase (YA,n) is related to the initial mole fraction in the liquid phase (XA,1) by:
YA,n1 − YA,n = (P°AP°B)n × XA,11 − XA,1

Where A is the more volatile component (P°A > P°B). As n → ∞, the vapour becomes enriched with the more volatile component (YA,n → 1).

21. Composition of Vapour Phase

  • When a liquid mixture evaporates, the composition of the first drop of vapour formed (YA, YB) relates to the liquid composition (XA, XB) as:
YA1 − YA = P°AP°B ∙ XA1 − XA

22. Composition of Last Drop of Liquid

  • As evaporation proceeds, the composition of the remaining liquid changes. For the last drop of liquid remaining (X'A, X'B), the vapour pressure is determined by its composition:
PT = P°A X'A + P°B X'B
  • The relation between the final vapour composition and the final liquid composition (X'A) follows the same principle:
YA1 − YA = P°AP°B ∙ X'A1 − X'A

23. Immiscible Volatile Liquids

  • When two liquids are immiscible, they do not mix.
  • The total vapor pressure is the sum of the pure vapor pressures of the individual components:
PT = P°A + P°B
  • The mole fraction of components in the vapor phase is given by:
yA = P°AP°A + P°B  ;  yB = P°BP°A + P°B

24. Vapor Phase Composition (Immiscible)

  • The ratio of moles of A (nA) to moles of B (nB) in the vapor phase is:
nAnB = P°AP°B
  • In terms of weights (WA, WB) and molar masses (MA, MB):
WAWB = P°AMAP°BMB

WA, WB = weight of A and B in the vapor phase.

25. Ideal Solutions

  • An ideal solution is formed when similar interactions are replaced by the same type of interaction.
  • Condition: A–B interaction = A–A and B–B interaction.
  • Example: Hexane + Heptane.
ΔHmix = 0
ΔVmix = 0
ΔGmix < 0
ΔSmix > 0
  • ΔHmix = 0 because energy required for bond breaking is equal to energy released during bond formation.
  • ΔVmix = 0 for A–A + B–B → A–B (e.g., 10 + 10 = 20).
  • ΔGmix < 0 since mixing occurs spontaneously.
  • ΔSmix > 0 because A–A + B–B → 2A–B increases randomness (symmetrical to unsymmetrical).

26. Raoult's Law Graph (Ideal)

  • Assuming A is more volatile: P°A > P°B.
PT = P°A XA + P°B XB
XAPT P°BP°A XA=00.2XA=1 XB=10.8XB=0

27. Examples of Ideal Solutions

  • Hexane + Heptane
  • Benzene + Toluene
  • Ethyl bromide + Ethyl Iodide
  • Chlorobenzene + Bromobenzene
Page 4
Non-Ideal Solutions (Positive Deviation)

28. Non-Ideal Solutions (Positive Deviation)

  • Characteristics:
ΔHmix > 0
ΔVmix > 0
ΔSmix > 0
ΔGmix < 0
  • Graph behavior: The total vapour pressure curve lies above the straight line ideal Raoult's law behavior.
  • PT > P°A XA + P°B XB

29. Examples of Positive Deviation

  • Ethanol + Cyclohexane
  • Ethanol + water or Methanol + water
  • Ethanol + Acetone
  • CCl4 + Benzene or CCl4 + Toluene
  • CCl4 + CHCl3
  • CCl4 + Methanol
  • Benzene + Acetone
  • Acetone + CS2
  • Acetaldehyde + CS2

30. Negative Deviation from Raoult's Law

  • Occurs when weaker solvent-solvent (A-A, B-B) interactions are replaced by stronger solute-solvent (A-B) interactions.
  • Condition: Interaction A-B > A-A and B-B interactions.
  • PT < PA°xA + PB°xB.
ΔHmix < 0 (Exothermic)
ΔVmix < 0
ΔSmix > 0
ΔGmix < 0

31. Graph: Negative Deviation

  • The total vapour pressure (PT) curve lies below the ideal line (dashed).
XAPT PB°PA° 01 0.2

32. Examples of Negative Deviation

  • CHCl3 + CH3COCH3 (Acetone)
  • CHCl3 + C6H6 (Benzene)
  • CHCl3 + C2H5OC2H5 (Diethylether)
  • H2O + HCl
  • H2O + HNO3
  • CH3COCH3 + Aniline
  • Phenol + Aniline

33. Chloroform and Acetone Mixture

  • Components: CHCl3 (Dipole-Dipole) and CH3COCH3 (Dipole-Dipole).
  • The mixture forms Hydrogen Bonds (H.B).
  • Interaction strength: H.B > Dipole-Dipole.

34. Non-Ideal Solutions

  • The main reason for deviation from ideal behavior in non-ideal solutions is the difference in intermolecular forces between the solute and solvent (A-B interaction) compared to pure components (A-A and B-B interactions).
  • Positive Deviation : Occurs when the A-B attraction is weaker than the A-A and B-B attractions (A-B attraction < A-A, B-B).
  • Negative Deviation : Occurs when the A-B attraction is stronger than the A-A and B-B attractions (A-B attraction > A-A, B-B).

35. Pressure-Composition Curves for Non-Ideal Solutions

  • Ideal Solution: Shows a linear or standard convex curve connecting P°B and P°A.
  • Negative Deviation:
    • Weak (-)ve deviation: Curve lies slightly below the ideal dashed line.
    • Strong (-)ve deviation: Curve lies significantly below the ideal dashed line.
  • Positive Deviation:
    • Weak (+)ve deviation: Curve lies above the ideal dashed line.
    • Strong (+)ve deviation: Curve exhibits a peak where XA = YA.

36. Azeotropic Mixture

  • A binary solution containing two volatile liquids where the liquid and vapour phases have the same composition.
  • A boiling liquid mixture at the azeotropic composition produces vapour with the exact same composition as the liquid.
  • Distillation cannot separate the two liquids in an azeotropic mixture because the condensate retains the same composition as the original mixture.

37. Minimum Boiling Azeotrope

  • These azeotropes are formed by solutions showing large positive deviation from Raoult's Law.
  • The boiling point of the solution is lower than the boiling point of either pure component.
  • Example shown: Water and Ethanol. The curve shows a maximum vapour pressure point at approximately 0.95 ethanol and 0.05 water.

38. Maximum Boiling Azeotrope

  • These azeotropes are formed by solutions which show negative deviation from Raoult's Law.
  • The boiling point of the azeotropic mixture is more than that of either individual component.
Page 5
Colligative Properties & Ebullioscopic Constant (Kb)

39. Colligative Properties

  • Properties of a solution that depend only on the number of solute particles and not on their nature.
  • Conditions:
    • Solute must be non-volatile.
    • Solution must be dilute.

40. Relative Lowering of Vapour Pressure (RLVP)

  • Vapour pressure of solution (Ps) = P° × Xsolvent
  • Relative Lowering of Vapour Pressure (RLVP) formula:
RLVP = P° − PsP° = Xsolute = nsolutensolute + nsolvent = WsoluteMsoluteWsoluteMsolute + WsolventMsolvent

Where: P° = v.p. of pure solvent, Ps = v.p. of solution, W = weight, M = molar mass.

Alternative Form :

  • Dividing by Xsolvent gives an exact form (useful when Ps is given): P° − PsPs = nsolutensolvent = Wsolute / MsoluteWsolvent / Msolvent

41. Comparison of RLVP

  • RLVP ∝ Xsolute.
  • Since Xsolute = nsolutensolute + nsolvent, for a fixed mass of solvent and equal masses of solutes, a smaller molar mass (Msolute) results in more moles (nsolute) and thus a higher RLVP.

42. Ostwald–Walker Experiment

  • This experiment is used to determine the Relative Lowering of Vapour Pressure (RLVP).
  • Dry air is passed sequentially through a solution, then pure solvent, and finally a U-tube containing an anhydrous dehydrating agent (CaCl2).
RLVP = P° − PsP° = W2W1 + W2

W1 = weight loss in solution container ; W2 = weight loss in solvent container.
Ps = Vapour pressure of solution ; P° = Vapour pressure of pure solvent.

43. Elevation in Boiling Point

  • Vapour pressure of a solution containing a non-volatile solute is lower than that of the pure solvent.
  • To reach atmospheric pressure (1 atm), the solution requires more heat, thus increasing its boiling point.
  • This elevation is a colligative property.
TV.P 1 atm TbTb'Tb''

Tb = B.P of solvent ; Tb' = B.P of solution 1 ; Tb'' = B.P of solution 2.

44. Ebullioscopic Constant (Kb)

  • The boiling point elevation (ΔTb) is proportional to molality (m):
ΔTb = Kb ⋅ m
  • Kb (Ebullioscopic Constant / Molal boiling point elevation constant) is a property of the solvent only.
  • Relationship for Kb:
Kb = M ⋅ R ⋅ Tb21000 ⋅ ΔHvap

M = molar mass of solvent ; Tb = boiling point of solvent ; ΔHvap = heat of vaporisation of solvent.

  • Since Kb ∝ Tb2ΔHvap:
  • (ΔHvap)H2O ≫ (ΔHvap)Benzene (due to H-bonding vs Vander Waal forces).
  • Consequently, (Kb)H2O < (Kb)Benzene.

45. Depression in Freezing Point

  • Freezing point: The temperature at which the first crystal of ice separates and the solid phase is in dynamic equilibrium with the liquid phase.
  • Condition: Freezing point is the temperature where the vapour pressure of the liquid phase equals the vapour pressure of the solid phase.
  • Only the solvent freezes.

46. Depression in Freezing Point

  • Freezing point itself is not a colligative property; however, the depression in freezing point (ΔTF) is a colligative property.
  • Graphically, the freezing point corresponds to the temperature where the vapour pressure of the liquid solvent/solution intersects with the vapour pressure of the solid (frozen) phase.
ΔTF = (TF)solvent − (TF)solution
  • As solute concentration increases, the vapour pressure of the solution decreases, leading to a shift in the freezing point intersection to lower temperatures (TF° > TF' > TF'').
  • Entropy relationships at equilibrium (solvent ⇌ ice): ΔG = ΔH − TΔS. At equilibrium ΔG = 0, so TF = ΔHfusΔS.
  • Since TF ∝ 1ΔS and ΔS1 < ΔS2 < ΔS3 (entropy differences between solution phases and ice), it follows that (TF)solvent > (TF)soln1 > (TF)soln2.
Page 6
Cryoscopic Constant (KF) & Osmotic Pressure (π)

47. Cryoscopic Constant (KF)

  • The depression in freezing point is directly proportional to the molality (m) of the solution:
ΔTF = KF ⋅ m
  • KF is the cryoscopic constant or molal depression in freezing point constant. It is a characteristic property of the solvent only.
  • Formula for KF:
KF = M R (TF)21000 ΔHfus

Where M = molar mass of solvent, R = gas constant, TF = freezing point of solvent, ΔHfus = heat of fusion of solvent.

48. Concept of Ice Separation

  • For the freezing process: ΔTF = KF ⋅ m.
  • As molality (m) increases, the depression in freezing point (ΔTF) increases, causing the freezing point (F.P.) of the solution to decrease.
  • Key Concept: During the cooling of a solution, the temperature until which the solution is being cooled is considered the freezing point of the remaining liquid solution.

49. Molal Depression and Elevation Constants

Kb = M R Tb2ΔHvap = M R TbΔSvap
Kf = M R Tf2ΔHfus = M R TfΔSfus

where M = molar mass of solvent (kg), R = gas constant, Tb = boiling point, Tf = freezing point, ΔH = enthalpy of vaporization/fusion, ΔS = entropy of vaporization/fusion.

  • Derived from ΔG = ΔH - TΔS. At freezing point (F.P.), ΔG = 0, so ΔHfus = TfΔSfus ⇒ ΔSfus = ΔHfusTf.

50. Osmotic Pressure

  • Semipermeable membrane: Artificial (Cu2Fe(CN)6) or Natural (animal bladder).
  • Osmosis: Movement of solvent molecules through a semipermeable membrane.
    • In terms of solvent concentration: High solvent conc. → Low solvent conc.
    • In terms of solution concentration: Low solute conc. → High solute conc.
[DIAGRAM: Container divided by a membrane. Left: 'Solvent'. Right: 'Solution' containing solute particles.]

51. Osmotic Pressure (π)

  • Osmotic pressure is the extra pressure applied on the more concentrated solution side to stop the process of osmosis.
  • It can also be measured by the hydrostatic pressure generated by the height (h) of the liquid column.
π = hρg

where h = height of liquid column, ρ = density, g = acceleration due to gravity.

  • At equilibrium: (rate)solvent → solution = (rate)solution → solvent.

52. Reverse Osmosis

  • When the pressure applied on the solution side (π') is greater than the osmotic pressure (π), the solvent flows from the solution side to the solvent side.
π' > π (Reverse Osmosis)

53. Calculation of Osmotic Pressure

π = CRT
  • C = concentration (molarity), R = gas constant, T = temperature.
  • Depending on units:
    • If π is in atm: R = 0.082 atm L / K-mol
    • If π is in Pa or N/m²: R = 8.314 J / K-mol
  • Standard conversion: 1 atm = 1.013 × 105 N/m².

54. Hypertonic, Hypotonic and Isotonic Solutions

  • For two solutions (1 and 2) with concentrations C1, C2 and temperatures T1, T2:
Isotonic: πsoln 1 = πsoln 2 ⇒ C1T1 = C2T2
If T1 = T2, then C1 = C2
  • If πsoln 1 > πsoln 2:
    • Soln 1 is Hypertonic.
    • Soln 2 is Hypotonic.
    • At the same temperature: C1 > C2.

55. Resultant Osmotic Pressure of Mixed Solutions

  • When two solutions of the same substance with different osmotic pressures (π1, π2) and volumes (V1, V2) are mixed:
πnet = π1V1 + π2V2V1 + V2

56. Van't Hoff Factor (i)

  • Due to association or dissociation of solute particles, the observed colligative property differs from the theoretical (calculated) colligative property.
i = Observed Colligative PropertyTheoretical Colligative Property
  • For any solute:
    i = McalMobs
Page 7
Colligative Properties with Van't Hoff Factor

57. Colligative Properties with Van't Hoff Factor

  • i) ΔTb = i Kb m
  • ii) ΔTf = i Kf m
  • iii) π = i CRT
  • iv) RLVP = i nsolutei nsolute + nsolvent ≈ i nsolutensolvent

58. Dissociation

  • Reaction: An ⇌ nA
  • If α is the degree of dissociation:
i = 1 + (n − 1)α

n = number of particles the solute dissociates into.

59. Association

  • Reaction: nA ⇌ An
  • If β is the degree of association:
i = 1 − (1 − 1n)β

n = number of particles associating together.

  • Example of association: CH3COOH (non-polar solvent).
  • Reaction: 2CH3COOH ⇌ (CH3COOH)2
  • [DIAGRAM: Dimerization of acetic acid via two hydrogen bonds between carboxylic acid groups.]

60. Non-Electrolytes

  • For solutes like glucose, sucrose, or urea:
i = 1
  • Here, (C.P.)calculated = (C.P.)observed.

61. Vapour Pressure of Electrolyte Solutions

  • For an aqueous solution of an electrolyte, the lowering of vapour pressure is calculated using the effective number of particles.
P° - PsP° = i × χsolute

Where P° is vapour pressure of pure solvent, Ps is vapour pressure of solution, and χsolute is mole fraction of solute.

62. Relative Lowering of Vapour Pressure & Solubility

  • For a salt with solubility S, dissolving as: XY₂₍ₛ₎ ⇌ X₍ₐq₎ + 2Y₍ₐq₎
  • Number of solute particles formed = S + 2S = 3S
P° − PₛPₛ = nsolutensolvent = 3Snsolvent

where S = solubility of the salt

63. Quick Formula Revision

QuantityFormula
Clausius–Clapeyronlog(P2/P1) = (ΔHvap/2.303R)(1/T1 − 1/T2)
Raoult's law (non-volatile solute)Ps = P° Xsolvent
RLVP(P° − Ps)/P° = Xsolute ; (P° − Ps)/Ps = nsolute/nsolvent
Two volatile liquidsPT = P°AXA + P°BXB = P°B + XA(P°A − P°B)
Vapour compositionYA = P°AXA/PT ; YA/(1 − YA) = (P°A/P°B) XA/(1 − XA)
PT from vapour composition1/PT = YA/P°A + YB/P°B
n-stage distillationYA,n/(1 − YA,n) = (P°A/P°B)n XA,1/(1 − XA,1)
Immiscible liquidsPT = P°A + P°B ; WA/WB = P°AMA/(P°BMB)
Ostwald–Walker(P° − Ps)/P° = W2/(W1 + W2)
Elevation in B.P.ΔTb = iKbm ; Kb = MRTb2/(1000 ΔHvap) (M in g mol−1)
Depression in F.P.ΔTf = iKfm ; Kf = MRTf2/(1000 ΔHfus)
Osmotic pressureπ = iCRT ; π = hρg ; mixing: π = (π1V1 + π2V2)/(V1 + V2)
van't Hoff factori = observed C.P./calculated C.P. = Mnormal/Mobserved
Dissociation / associationi = 1 + (n − 1)α ; i = 1 − (1 − 1/n)β

Key Points :

  • V.P. of a pure liquid depends only on temperature and rises exponentially with T; stronger intermolecular forces mean lower V.P.
  • In the vapour, the more volatile component is always richer than in the liquid; repeated distillation enriches it further.
  • Ideal solution: ΔHmix = 0, ΔVmix = 0, ΔSmix > 0, ΔGmix < 0 (e.g. benzene + toluene, hexane + heptane).
  • Positive deviation (A–B weaker): ΔHmix > 0, ΔVmix > 0, higher V.P., minimum boiling azeotrope (ethanol + water).
  • Negative deviation (A–B stronger): ΔHmix < 0, ΔVmix < 0, lower V.P., maximum boiling azeotrope (HNO3 + water, CHCl3 + acetone).
  • Azeotropes boil with the same composition in liquid and vapour, so they cannot be separated by fractional distillation.
  • Colligative properties depend only on the number of solute particles; Kb and Kf depend only on the solvent.
  • On cooling a solution below its freezing point, only pure solvent separates as ice; the remaining solution freezes at the temperature it is cooled to.
  • Isotonic solutions have equal iC at the same temperature; solvent flows from lower to higher iC (reverse osmosis when applied pressure > π).
  • Electrolytes have i > 1, non-electrolytes i = 1, and acids that dimerise in benzene (acetic, benzoic) have i < 1.