Solutions
Concentration Terms, Henry's Law, Raoult's Law, Ideal & Non-Ideal Solutions, Colligative Properties, Osmosis & van't Hoff Factor
Topic 1.1 – 1.2: Types of Solutions & Expressing Concentration (Mass %, ppm, χ, M & m)
Look around you—almost nothing in everyday life is a pure substance! The brass in temple bells (copper + zinc), the air we breathe, the 35% ethylene glycol antifreeze keeping car radiators from freezing in Ladakh, and even the 1 ppm fluoride in drinking water that protects tooth enamel (while 1.5 ppm mottles teeth!) are all homogeneous mixtures called solutions.
A solution is a homogeneous mixture of two or more chemically non-reacting components whose composition and properties are uniform throughout. The component present in the largest quantity is the solvent (which sets the physical state of the solution), and the remaining components are solutes.
Nine Types of Binary Solutions (NCERT Table 1.1)
| Solution Type | Solute | Solvent | NCERT Example |
|---|---|---|---|
| Gaseous | Gas / Liquid / Solid | Gas | O2 + N2 (air) / Chloroform in N2 / Camphor in N2 |
| Liquid | Gas / Liquid / Solid | Liquid | O2 in water / Ethanol in water / Glucose in water |
| Solid | Gas / Liquid / Solid | Solid | H2 in Palladium / Mercury in Sodium (amalgam) / Cu in Gold |
Seven Quantitative Concentration Expressions
- Mass % (w/w): (Mass of component / Total mass of solution) × 100. Used industrially (e.g., commercial bleaching solution = 3.62% NaOCl).
- Volume % (V/V): (Volume of component / Total volume of solution) × 100. E.g., 35% (V/V) ethylene glycol antifreeze lowers water's freezing point to 255.4 K (−17.6 °C).
- Mass by Volume % (w/V): Mass of solute (g) dissolved in 100 mL of solution (used in medicine & pharmacy).
- Parts per million (ppm): (Parts of component / Total parts of all components) × 106. Ideal for trace pollutants & dissolved O2 in sea water (6 × 10−3 g in 1030 g = 5.8 ppm).
Molarity (M): Moles of solute / Volume of solution in L (mol L-1)
Molality (m): Moles of solute / Mass of solvent in kg (mol kg-1)
Density Shortcut: M = (% w/w × d × 10) / Msolute | m = (1000 × M) / (1000 × d − M × Msolute)
• Ex 1.1: 20% (w/w) aqueous ethylene glycol (C2H6O2, M = 62): in 100 g solution, nglycol = 20/62 = 0.322 mol, nwater = 80/18 = 4.444 mol → χglycol = 0.322 / 4.766 = 0.068, χwater = 1 − 0.068 = 0.932.
• Ex 1.2: 5 g NaOH (M = 40) in 450 mL solution → n = 5/40 = 0.125 mol → M = 0.125 / 0.450 = 0.278 mol L-1.
• Ex 1.3: 2.5 g CH3COOH (M = 60) in 75 g benzene → n = 2.5/60 = 0.0417 mol → m = 0.0417 / 0.075 = 0.556 mol kg-1.
Topic 1.3: Solubility of Solids & Gases in Liquids and Henry's Law
Open a chilled bottle of soda and hear that crisp psshhh! followed by hundreds of fizzing CO2 bubbles racing upward. Why did the gas stay peacefully hidden inside the sealed bottle, and why does warm soda go flat much faster than cold soda? One simple equation by William Henry answers soda factories, scuba divers, mountain climbers, and river fish alike!
1.3.1 Solubility of a Solid in a Liquid
- "Like Dissolves Like": Polar/ionic solutes (NaCl, sugar) dissolve in polar solvents (water); non-polar solutes (naphthalene, anthracene) dissolve in non-polar solvents (benzene).
- Dynamic Equilibrium (Saturated Solution): Rate of dissolution equals rate of crystallisation.
- Temperature Effect (Le Chatelier): Endothermic dissolution (ΔsolH > 0) → solubility increases with T. Exothermic dissolution (ΔsolH < 0) → solubility decreases with T.
- Pressure Effect: Negligible for solids/liquids because they are highly incompressible.
1.3.2 Solubility of a Gas in a Liquid & Henry's Law
Henry's Law: At constant temperature, the partial pressure of a gas in vapour phase (p) is directly proportional to the mole fraction of the gas (x) in the solution:
- Effect of Temperature: Gas dissolution is exothermic (similar to condensation). Thus KH increases with temperature and gas solubility decreases when heated—explaining why aquatic species are more comfortable in cold waters than warm waters.
- Soft Drinks & Soda Water: Bottled under high CO2 pressure to force more CO2 into solution.
- Scuba Divers & "Bends": Breathing high-pressure air underwater dissolves large amounts of N2 in blood. Ascending too fast drops pressure, releasing painful, life-threatening N2 bubbles that block capillaries (bends). Prevention: Tanks are diluted with Helium (11.7% He, 56.2% N2, 32.1% O2).
- High-Altitude Anoxia: Low atmospheric O2 partial pressure at high altitudes lowers blood O2, causing climbers to feel weak and unable to think clearly (anoxia).
• x(N2) = 0.987 / 76,480 = 1.29 × 10-5.
• 1 L water = 1000/18 = 55.5 mol → n / 55.5 = 1.29 × 10-5 → n = 7.16 × 10-4 mol = 0.716 mmol.
Topic 1.4: Vapour Pressure of Liquid Solutions & Raoult's Law
Picture a crowded exit door at a cinema hall. In pure water, every person standing at the exit is a water molecule ready to step outside (evaporate). Now sprinkle non-volatile sugar molecules along the doorway—they don't leave, they simply block half the doorway! Fewer water molecules escape per second, so vapour pressure drops!
1.4.1 Binary Solution of Two Volatile Liquids
Raoult's Law (1886): For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in solution.
ptotal = p1 + p2 = p1° + (p2° − p1°) x2 | Vapour Phase Composition: pi = yi · ptotal
- p1° and p2° are vapour pressures of pure components 1 and 2 at the same temperature.
- Key Rule: At equilibrium, the vapour phase is always richer in the more volatile component (higher p° gives larger yi).
1.4.2 Raoult's Law as a Special Case of Henry's Law
Comparing pi = pi° xi (Raoult) and p = KH x (Henry), both state that partial pressure of a volatile component is proportional to its liquid mole fraction—Raoult's law is simply the special case of Henry's law when KH = p1°.
1.4.3 Non-Volatile Solute in a Volatile Solvent
When a non-volatile solid (glucose, urea, sucrose) is added, solute particles occupy part of the liquid surface, reducing escaping solvent molecules. Thus psolution = p1 = p1° x1 < p1°—depending only on solute quantity, not chemical identity.
• n(CH2Cl2) = 40/85 = 0.47 mol; n(CHCl3) = 25.5/119.5 = 0.213 mol → Total = 0.683 mol.
• x(CH2Cl2) = 0.688, x(CHCl3) = 0.312.
• ptotal = 200 + (415 − 200) × 0.688 = 347.9 mm Hg.
• Partial pressures: p(CH2Cl2) = 0.688 × 415 = 285.5 mm Hg; p(CHCl3) = 0.312 × 200 = 62.4 mm Hg.
• Vapour fractions: y(CH2Cl2) = 285.5 / 347.9 = 0.82; y(CHCl3) = 0.18 (vapour is richer in more volatile CH2Cl2!).
Topic 1.5: Ideal & Non-Ideal Solutions, Deviations from Raoult's Law & Azeotropes
Think of mixing two friend groups at a college fest. If guests from A and B bond equally comfortably with each other as with their old friends, nothing heats up or stretches out—that's an Ideal Solution. If A and B dislike each other and push away, they try to escape the party faster (Positive Deviation). If A and B suddenly hug tightly via new hydrogen bonds, nobody wants to leave (Negative Deviation)!
| Property | Ideal Solution | Positive Deviation (+ve) | Negative Deviation (−ve) |
|---|---|---|---|
| Raoult's Law | Obeys across all compositions | ptotal > p1°x1 + p2°x2 | ptotal < p1°x1 + p2°x2 |
| Molecular Force | F(A–B) ≈ F(A–A), F(B–B) | F(A–B) weaker than A–A, B–B | F(A–B) stronger than A–A, B–B |
| ΔmixH & ΔmixV | ΔmixH = 0, ΔmixV = 0 | ΔmixH > 0 (endo), ΔmixV > 0 | ΔmixH < 0 (exo), ΔmixV < 0 |
| Azeotrope | None (separable by distillation) | Minimum boiling azeotrope | Maximum boiling azeotrope |
| NCERT Examples | n-hexane + n-heptane; benzene + toluene; bromoethane + chloroethane | Ethanol + acetone; CS2 + acetone; ethanol + water (95%) | Chloroform + acetone; phenol + aniline; HNO3 + water (68%) |
What are Azeotropes?
Azeotropes are binary liquid mixtures that have the exact same composition in liquid and vapour phases and boil at a constant temperature—making separation by fractional distillation impossible once reached.
- Minimum Boiling Azeotrope: Formed by solutions with large positive deviation. E.g., ~95% by volume ethanol + water obtained from sugar fermentation.
- Maximum Boiling Azeotrope: Formed by solutions with large negative deviation. E.g., 68% HNO3 + 32% water by mass (boils at 393.5 K).
Topic 1.6 – 1.6.1: Colligative Properties & Relative Lowering of Vapour Pressure (RLVP)
Whether you drop 1 mole of tiny urea molecules (60 g) or 1 mole of giant sucrose molecules (342 g) into 1 kg of water, water's vapour pressure drops by the exact same amount! Why? Because these properties behave like a bouncer counting heads at the door—they care ONLY about how many particles are inside, never who they are!
Properties of dilute solutions that depend solely on the number of solute particles relative to the total number of particles, and not on their chemical nature, are called Colligative Properties (Latin: co = together, ligare = bind):
- Relative lowering of vapour pressure (RLVP)
- Elevation of boiling point (ΔTb)
- Depression of freezing point (ΔTf)
- Osmotic pressure (Π)
1.6.1 Relative Lowering of Vapour Pressure (RLVP)
For a non-volatile solute (2) in volatile solvent (1), p1 = x1 p1°. Thus lowering Δp1 = p1° − p1 = p1°(1 − x1) = p1° x2. Dividing by p1°:
For dilute solutions (n2 ≪ n1): (p1° − p1) / p1° ≈ n2 / n1 = (w2 × M1) / (M2 × w1)
• (0.850 − 0.845) / 0.850 = (0.5 × 78) / (M2 × 39) → 0.005 / 0.850 = 1 / M2 → M2 = 170 g mol-1.
Topic 1.6.2 – 1.6.3: Elevation of Boiling Point (ΔT_b) & Depression of Freezing Point (ΔT_f)
Why do snow-plow trucks scatter salt on icy Himalayan highways, and why do motorists pour ethylene glycol into engine coolants? Adding solute lowers water's vapour pressure curve—simultaneously forcing the liquid to boil hotter (preventing boil-over!) and freeze colder (preventing radiator ice!). Two problems solved by one molecular physics trick!
1.6.2 Elevation of Boiling Point (Ebullioscopy)
A liquid boils when its vapour pressure reaches atmospheric pressure (1.013 bar). Because a non-volatile solute lowers vapour pressure at 373.15 K below 1.013 bar, the solution must be heated to a higher temperature Tb > Tb°.
Kb is the Boiling Point Elevation / Molal Elevation / Ebullioscopic Constant (unit: K kg mol-1; for water, Kb = 0.52 K kg mol-1).
1.6.3 Depression of Freezing Point (Cryoscopy)
Freezing occurs when the vapour pressure of liquid equals that of solid solvent. Because solution vapour pressure is lowered, it meets the solid curve at a lower temperature Tf < Tf°.
Thermodynamic Constants: Kf = (R · M1 · Tf2) / (1000 · ΔfusH) | Kb = (R · M1 · Tb2) / (1000 · ΔvapH)
| Solvent (NCERT Table 1.3) | b.p. (K) | Kb (K kg mol-1) | f.p. (K) | Kf (K kg mol-1) |
|---|---|---|---|---|
| Water | 373.15 | 0.52 | 273.0 | 1.86 |
| Benzene | 353.3 | 2.53 | 278.6 | 5.12 |
| Cyclohexane | 353.74 | 2.79 | 279.55 | 20.00 |
• Ex 1.7: 18 g glucose (0.1 mol) in 1 kg water → m = 0.1 → ΔTb = 0.52 × 0.1 = 0.052 K → Tb = 373.202 K.
• Ex 1.8: 1.80 g solute in 90 g benzene raises BP from 353.23 K to 354.11 K (ΔTb = 0.88 K) → M2 = (2.53 × 1.80 × 1000)/(0.88 × 90) = 58 g mol-1.
• Ex 1.9: 45 g ethylene glycol (0.73 mol) in 600 g water → m = 1.2 → ΔTf = 1.86 × 1.2 = 2.2 K → Tf = 270.95 K.
• Ex 1.10: 1.00 g solute in 50 g benzene lowers FP by 0.40 K → M2 = (5.12 × 1.00 × 1000)/(0.40 × 50) = 256 g mol-1.
Topic 1.6.4 – 1.6.5: Osmosis, Osmotic Pressure (Π = CRT) & Reverse Osmosis Desalination
Drop a limp, wilted rose stem into a glass of fresh water and it perks right back up. Drop raw mango pieces into concentrated brine and they shrivel into delicious Kerala uppumanga pickle! Why? Cell walls act as semipermeable membranes, letting water rush toward whichever side holds more dissolved salt or sugar!
1.6.4 Osmosis & Osmotic Pressure
- Semipermeable Membrane (SPM): Films with submicroscopic pores (parchment, pig's bladder, cellophane, cellulose acetate) that permit small solvent molecules to pass while blocking larger solute molecules/ions.
- Osmosis: Spontaneous net flow of solvent molecules through an SPM from pure solvent (or dilute solution) into a concentrated solution.
- Osmotic Pressure (Π): The excess hydrostatic pressure applied on the solution side just enough to stop osmosis.
Why Osmotic Pressure is BEST for Proteins, Polymers & Macromolecules:
- Measured around room temperature (biomolecules denature/decompose when heated for ΔTb!).
- Uses molarity (C) instead of molality, and gives large measurable pressure even for very dilute solutions of huge polymers.
Isotonic, Hypertonic & Hypotonic Solutions (Medical Biology)
- Isotonic: Two solutions with identical osmotic pressure (no net osmosis). Blood cell fluid is isotonic with 0.9% (w/v) NaCl (normal saline).
- Hypertonic (> 0.9% NaCl): Water flows OUT of blood cells → cells shrink. (Also preserves salted meat/candied fruit by dehydrating bacteria; excess dietary salt causes water retention / edema!)
- Hypotonic (< 0.9% NaCl): Water flows INTO blood cells → cells swell and may burst.
1.6.5 Reverse Osmosis (RO) & Desalination
When pressure greater than osmotic pressure (Pext > Π) is applied on the solution side, pure solvent flows out of the solution through a cellulose acetate SPM—used globally for desalination of sea water.
• M2 = (1.26 × 0.083 × 300) / (2.57 × 10-3 × 0.200) = 61,022 g mol-1.
Topic 1.7: Abnormal Molar Masses & van't Hoff Factor (i)
Imagine counting guests entering a hall: you expect 10 couples (10 pairs) to add 10 units to the crowd. Suddenly every couple splits up—giving 20 individuals! Or imagine 10 single cyclists pairing up on tandem bikes—leaving only 5 bikes on the road! Electrolytes like KCl split (dissociate) in water while carboxylic acids pair up (associate) in benzene—producing abnormal molar masses until van't Hoff fixed the count with factor i!
When a solute dissociates into ions (KCl → K+ + Cl−), particle count increases, colligative properties double, and experimentally calculated molar mass becomes lower than true molar mass (e.g. KCl appears ~37.25 instead of 74.5 g mol-1). When molecules associate (e.g., 2 CH3COOH ⇌ (CH3COOH)2 via H-bonding in benzene), particle count halves and observed molar mass becomes higher (double).
i = (Normal Molar Mass) / (Abnormal Molar Mass) = (Observed Colligative Property) / (Calculated Colligative Property)
i = (Total moles of particles AFTER association/dissociation) / (Moles BEFORE association/dissociation)
• Dissociation (1 → n ions, degree α): i = 1 + (n − 1)α (i > 1)
• Association (n molecules → 1, degree α): i = 1 − α(1 − 1/n) (i < 1)
Modified Colligative Equations & NCERT Table 1.4 Values
- RLVP: (p1° − p1) / p1° = i · (n2 / n1) | BP Elevation: ΔTb = i · Kb · m
- FP Depression: ΔTf = i · Kf · m | Osmotic Pressure: Π = i · (n2RT / V) = i · C R T
- At complete dissociation (dilute limit): NaCl, KCl, MgSO4 → i = 2; CaCl2, K2SO4 → i = 3; K4[Fe(CN)6] → i = 5. (At higher concentrations like 0.1 m, interionic attraction lowers i: NaCl = 1.87, MgSO4 = 1.21).
• Ex 1.12 (Association): 2 g benzoic acid (Mnormal = 122) in 25 g benzene gives ΔTf = 1.62 K (Kf = 4.9). Observed M2 = (4.9 × 2 × 1000)/(25 × 1.62) = 241.98 g mol-1.
i = 122 / 241.98 = 0.504. For dimerisation (1 − x/2 = i): x/2 = 1 − 0.504 = 0.496 → x = 0.992 (99.2% associated!).
• Ex 1.13 (Weak Electrolyte Dissociation): 0.6 mL CH3COOH (d = 1.06 g/mL → 0.0106 mol) in 1 L water gives ΔTf,obs = 0.0205 K (ΔTf,calc = 1.86 × 0.0106 = 0.0197 K) → i = 0.0205 / 0.0197 = 1.041 → α = i − 1 = 0.041 → Ka = cα2/(1 − α) = 1.86 × 10-5.
Solved Previous Year Questions (Kerala DHSE & NCERT)
Q1 (Kerala HSE March (Repeated Favorite) • 1 Marks): Why is molality preferred over molarity in expressing the concentration of a solution when temperature varies?
Answer: Molality (m) depends on the mass of the solvent, which is independent of temperature. On the other hand, Molarity (M) depends on the volume of solution, which expands or contracts with temperature.
Q2 (Kerala HSE SAY • 1 Marks): Aquatic species are more comfortable in cold waters rather than in warm waters. Give reason.
Answer: Dissolution of oxygen gas in water is exothermic, so KH increases with increase in temperature. At lower temperature (cold water), KH is smaller and hence the solubility of dissolved O2 is higher, making aquatic life more comfortable.
Q3 (Kerala HSE Model • 1 Marks): Which colligative property is best suited for determining the molar masses of biomolecules such as proteins and polymers? Give one reason.
Answer: Osmotic pressure (Π). Reason: It is measured around room temperature (so fragile proteins do not denature) and gives a relatively large, easily measurable magnitude even for very dilute solutions of macromolecules.
Q4 (Kerala HSE March • 2 Marks): State Henry's law. Write any two applications of Henry's law.
Statement: At constant temperature, the partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (x) in the solution: p = KH x.
Applications (any two):
1. In carbonated soft drinks/soda water, bottles are sealed under high CO2 pressure to increase gas solubility.
2. Scuba diving tanks are diluted with Helium (11.7% He, 56.2% N2, 32.1% O2) to prevent bends (N2 bubbles in blood during ascent).
3. Explains anoxia (low blood O2) at high altitudes due to low partial pressure of oxygen.
Q5 (Kerala HSE March • 2 Marks): Distinguish between ideal and non-ideal solutions with respect to Raoult's law, ΔmixH, and ΔmixV. Give one example for each.
Ideal Solutions: Obey Raoult's law over the entire concentration range; ΔmixH = 0 and ΔmixV = 0 (A–B interactions ≈ A–A, B–B). Example: Benzene + Toluene (or n-hexane + n-heptane).
Non-Ideal Solutions: Do NOT obey Raoult's law over the entire concentration range; ΔmixH ≠ 0 and ΔmixV ≠ 0. Example: Ethanol + Acetone (+ve deviation) or Chloroform + Acetone (−ve deviation).
Q6 (Kerala HSE SAY • 2 Marks): What are azeotropes? Distinguish between minimum boiling azeotropes and maximum boiling azeotropes with one example each.
Azeotropes: Binary liquid mixtures having the same composition in liquid and vapour phases that boil at a constant temperature (cannot be separated by fractional distillation).
1. Minimum boiling azeotrope: Formed by solutions showing large positive deviation from Raoult's law (e.g., ~95% ethanol + water).
2. Maximum boiling azeotrope: Formed by solutions showing large negative deviation from Raoult's law (e.g., 68% HNO3 + 32% water).
Q7 (Kerala HSE March • 2 Marks): Define osmosis and isotonic solutions. What happens when blood cells are placed in (i) > 0.9% (w/v) NaCl and (ii) < 0.9% (w/v) NaCl?
Osmosis: Net spontaneous flow of solvent molecules through a semipermeable membrane from pure solvent (or dilute solution) to concentrated solution.
Isotonic solutions: Two solutions having the same osmotic pressure at a given temperature (no net osmosis).
(i) In > 0.9% NaCl (hypertonic): water flows out of cells and they shrink.
(ii) In < 0.9% NaCl (hypotonic): water flows into cells and they swell.
Q8 (Kerala HSE Model • 2 Marks): What is reverse osmosis? Mention one important application and name the semipermeable membrane commonly used.
Reverse Osmosis (RO): When an external pressure greater than the osmotic pressure (P > Π) is applied to the solution side, pure solvent flows out of the solution through the semipermeable membrane.
Application: Desalination of sea water to produce potable water.
Membrane: A thin film of cellulose acetate placed over a suitable support.
Q9 (Kerala HSE March (NCERT Ex 1.8) • 3 Marks): The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (Kb for benzene = 2.53 K kg mol-1)
Given: w2 = 1.80 g, w1 = 90 g, Kb = 2.53 K kg mol-1.
Elevation in boiling point: ΔTb = 354.11 − 353.23 = 0.88 K.
Formula: M2 = (Kb × w2 × 1000) / (ΔTb × w1)
M2 = (2.53 × 1.80 × 1000) / (0.88 × 90) = 4554 / 79.2 = 57.5 ≈ 58 g mol-1.
Q10 (Kerala HSE SAY (NCERT Ex 1.9) • 3 Marks): 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution. (Kf for water = 1.86 K kg mol-1; Molar mass of C2H6O2 = 62 g mol-1)
Moles of ethylene glycol = 45 / 62 = 0.726 ≈ 0.73 mol.
Mass of water = 600 g = 0.60 kg → Molality m = 0.73 / 0.60 = 1.2 mol kg-1 (or 1.21 m).
(a) ΔTf = Kf × m = 1.86 × 1.2 = 2.2 K (2.25 K unrounded).
(b) Freezing point Tf = 273.15 − 2.2 = 270.95 K (270.9 K).
Q11 (Kerala HSE / NCERT Ex 1.11 • 3 Marks): 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein. (R = 0.083 L bar mol-1 K-1)
Given: w2 = 1.26 g, V = 200 cm3 = 0.200 L, T = 300 K, Π = 2.57 × 10-3 bar, R = 0.083 L bar mol-1 K-1.
M2 = (w2 × R × T) / (Π × V) = (1.26 × 0.083 × 300) / (2.57 × 10-3 × 0.200) = 61,022 g mol-1 (or 61,039 g mol-1).
Q12 (Kerala HSE March (NCERT Ex 1.12) • 4 Marks): Define van't Hoff factor (i). What is its value for complete dissociation of K2SO4? 2 g of benzoic acid (C6H5COOH, M = 122 g mol-1) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K (Kf = 4.9 K kg mol-1). Calculate the percentage association of the acid if it forms a dimer in solution.
van't Hoff factor (i): Ratio of normal molar mass to abnormal (observed) molar mass, or ratio of observed colligative property to calculated colligative property. For complete dissociation of K2SO4 (→ 2K+ + SO42-), i = 3.
Numerical:
Mabnormal = (4.9 × 2 × 1000) / (25 × 1.62) = 241.98 g mol-1.
i = Mnormal / Mabnormal = 122 / 241.98 = 0.504.
For dimerisation 2 C6H5COOH ⇌ (C6H5COOH)2: i = 1 − α/2 → α/2 = 1 − 0.504 = 0.496 → α = 0.992 → Percentage association = 99.2%.