Solutions — Raoult's Law & Colligative Properties
Most numericals in this chapter rest on one idea: some properties of a solution depend only on how many solute particles there are, not on what they are. Count the particles correctly, including what happens when they dissociate or associate, and the formulas do the rest.
What you will be able to do
- Convert between molarity, molality and mole fraction, and know which change with temperature
- Use Henry’s law, and Raoult’s law for both liquid and vapour composition
- Classify a solution as ideal or showing positive or negative deviation, and explain azeotropes
- Solve all four colligative-property numericals, including finding molar mass
- Use the van’t Hoff factor for electrolytes and associating solutes
The big picture
One cause, four effects.
In one line: A non-volatile solute lowers the solvent’s vapour pressure; that one change raises the boiling point, lowers the freezing point and creates osmotic pressure — all in proportion to the number of particles.
Think of it like this: Picture the surface of the liquid as a doorway. Solvent molecules escape through it to make vapour. Solute particles stand in the doorway and block some of the exits, so fewer solvent molecules escape. It doesn’t matter whether the blocker is glucose or urea; only how many blockers there are.
Key idea: “Colligative” means “depending on the collection”: the number of particles, not their identity. That is why NaCl (two ions per formula unit) has nearly twice the effect of glucose at the same molality.
Concentration terms
Pick the right unit before you pick the formula.
| Term | Definition | Changes with temperature? |
|---|---|---|
| Molarity (M) | mol solute per litre of solution | Yes (volume expands) |
| Molality (m) | mol solute per kg of solvent | No |
| Mole fraction (x) | mol of a component ÷ total mol | No |
| Mass percent | g solute per 100 g solution | No |
| ppm | g solute per 10⁶ g solution | No |
Key idea: Molality and mole fraction depend only on masses, so they don’t change with temperature. That is why ΔTb and ΔTf formulas use molality.
Solved example 01 — Molality and mole fraction
5.85 g NaCl is dissolved in 500 g water. Find the molality. Separately, find the mole fraction of solute in a 1 molal aqueous solution.
- mol NaCl = 5.85/58.5 = 0.100; molality = 0.100/0.500 kg = 0.200 mol kg⁻¹.
- 1 molal: 1 mol solute per 1000 g water = 55.5 mol water.
- x(solute) = 1/(1 + 55.5) = 0.0177.
Answer: 0.200 mol kg⁻¹; x(solute) ≈ 0.0177.
Henry’s law
How much gas dissolves in a liquid.
p = K(H) × xp = partial pressure of the gas above the solution; x = its mole fraction in solution
- Higher K(H) → lower solubility. K(H) rises with temperature, so gases are less soluble in warm water (fish suffer in warm ponds).
- Soft drinks are bottled under high CO₂ pressure; opening the bottle drops the pressure and CO₂ comes out.
- Divers breathing compressed air dissolve extra N₂ in their blood; surfacing too fast releases bubbles (“the bends”). Diving tanks use helium-diluted air.
- At high altitude, the low partial pressure of O₂ means less dissolved O₂ (anoxia).
Solved example 02 — Gas dissolved in water
K(H) for N₂ in water at 298 K is 76.48 kbar. Air above the water has p(N₂) = 0.80 bar. How many moles of N₂ dissolve in 1 L of water?
- x(N₂) = p/K(H) = 0.80/76,480 = 1.05 × 10⁻⁵.
- 1 L water = 55.5 mol. Since x is tiny, mol N₂ ≈ x × 55.5.
Answer: About 5.8 × 10⁻⁴ mol N₂.
Raoult’s law
Vapour pressure of a solution, for volatile and non-volatile solutes.
p(A) = x(A) × p°(A)p(total) = x(A)p°(A) + x(B)p°(B)
Each volatile component contributes in proportion to its mole fraction
Vapour composition: the mole fraction of A in the vapour is y(A) = p(A)/p(total). The vapour is always richer in the more volatile component — the basis of fractional distillation.
(p° − p) / p° = x(solute)Non-volatile solute: relative lowering of vapour pressure = mole fraction of solute
Solved example 03 — Two volatile liquids
26 g heptane (p° = 105.2 kPa, M = 100) is mixed with 35 g octane (p° = 46.8 kPa, M = 114). Find the vapour pressure and the mole fraction of heptane in the vapour.
- n(heptane) = 0.260; n(octane) = 0.307; x(heptane) = 0.260/0.567 = 0.459.
- p = 0.459 × 105.2 + 0.541 × 46.8 = 48.3 + 25.3 = 73.6 kPa.
- y(heptane) = 48.3/73.6.
Answer: p ≈ 73.6 kPa; y(heptane) ≈ 0.66 — the vapour is richer in the more volatile heptane.
Solved example 04 — Molar mass from RLVP
10 g of a non-volatile solute dissolved in 90 g water lowers the vapour pressure by 2%. Find its molar mass.
- (p° − p)/p° = 0.02 = n₂/(n₁ + n₂), with n₁ = 90/18 = 5.0.
- n₂ = 0.02 × 5.0/0.98 = 0.102 mol.
- M = 10/0.102.
Answer: M ≈ 98 g mol⁻¹.
Check yourself
- C1. Which unit of concentration changes when the solution is warmed?
- C2. Why is the vapour above a benzene–toluene mixture richer in benzene?
Answers at the end.
Ideal and non-ideal solutions
When Raoult’s law holds, and what happens when it doesn’t.
| Ideal | Positive deviation | Negative deviation | |
|---|---|---|---|
| A–B attraction vs A–A, B–B | Equal | Weaker | Stronger |
| Vapour pressure | Obeys Raoult | Higher than predicted | Lower than predicted |
| ΔH(mix) | 0 | Positive (absorbs heat) | Negative (releases heat) |
| ΔV(mix) | 0 | Positive (expands) | Negative (contracts) |
| Examples | Benzene + toluene; n-hexane + n-heptane | Ethanol + acetone; CS₂ + acetone; ethanol + water | Chloroform + acetone; HNO₃ + water; phenol + aniline |
| Azeotrope type | None | Minimum-boiling | Maximum-boiling |
Think of it like this: Positive deviation: the two liquids would rather be with their own kind, so molecules escape more easily (higher vapour pressure). Negative deviation: they like each other more than themselves — chloroform’s H bonds to acetone’s oxygen — so they hold on and escape less.
Azeotropes boil at constant composition, so fractional distillation can’t separate them. About 95% ethanol + water is a minimum-boiling azeotrope (positive deviation); about 68% HNO₃ + water is maximum-boiling (negative deviation).
The four colligative properties
Same cause, four measurements.
Relative lowering of VP = x(solute)ΔTb = Kb × m
ΔTf = Kf × m
π = CRT
m = molality; C = molarity; R = 0.083 L bar K⁻¹ mol⁻¹
| Solvent | Kb (K kg mol⁻¹) | Kf (K kg mol⁻¹) |
|---|---|---|
| Water | 0.52 | 1.86 |
| Benzene | 2.53 | 5.12 |
| Camphor | — | ≈ 40 (used for small samples) |
Molar mass from ΔTb (use Kb) or ΔTf (use Kf); masses in grams
Solved example 05 — Freezing point
1.0 g urea (M = 60) is dissolved in 100 g water. Find the freezing point (Kf = 1.86).
- m = (1.0/60)/0.100 = 0.167 mol kg⁻¹.
- ΔTf = 1.86 × 0.167 = 0.31 K.
Answer: Freezes at about −0.31 °C.
Solved example 06 — Molar mass from boiling point
1.8 g of a non-volatile solute in 100 g benzene raises the boiling point by 0.36 K (Kb = 2.53). Find M.
- M = 2.53 × 1.8 × 1000/(0.36 × 100).
Answer: M ≈ 126.5 g mol⁻¹.
Osmosis and osmotic pressure
The colligative property that works best for big molecules.
Osmosis is the flow of solvent through a semipermeable membrane from the dilute side to the concentrated side. Osmotic pressure π is the pressure needed to stop it.
- Isotonic solutions have equal π (0.9% NaCl matches blood cells).
- Hypertonic outside → water leaves the cell → it shrinks (plasmolysis). Hypotonic outside → water enters → it swells and may burst.
- Reverse osmosis: apply pressure above π to push water out of salt water — desalination.
- Why π for polymers and proteins? Even a very dilute solution gives a large, measurable π, it’s measured at room temperature (no denaturing), and it uses molarity directly.
Solved example 07 — Molar mass of a protein
1.26 g of a protein in 200 mL of solution has π = 2.57 × 10⁻³ bar at 300 K. Find M.
- π = (w/MV)RT, so M = wRT/(πV).
- M = 1.26 × 0.083 × 300/(2.57 × 10⁻³ × 0.200).
Answer: M ≈ 61,000 g mol⁻¹. (ΔTf for this solution would be only about 10⁻⁴ K — too small to measure.)
The van’t Hoff factor
When the number of particles isn’t the number of formula units.
i = observed colligative property / calculated (no dissociation)ΔTf = i Kf m · ΔTb = i Kb m · π = iCRT
| Behaviour | i | Example |
|---|---|---|
| Non-electrolyte | 1 | Glucose, urea, sucrose |
| Dissociates into n ions | 1 < i ≤ n | NaCl → 2; CaCl₂ → 3; K₄[Fe(CN)₆] → 5 |
| Associates (n molecules → 1) | < 1 | Ethanoic or benzoic acid dimerises in benzene: i → 0.5 |
Association: α = (1 − i)/(1 − 1/n)
Solved example 08 — Degree of dissociation
0.10 m NaCl freezes at −0.348 °C (Kf = 1.86). Find α.
- Expected ΔTf = 1.86 × 0.10 = 0.186 K.
- i = 0.348/0.186 = 1.87.
- α = (1.87 − 1)/(2 − 1).
Answer: α ≈ 0.87 (87% dissociated).
Solved example 09 — Association in benzene
2.0 g benzoic acid (M = 122) in 25 g benzene lowers the freezing point by 1.62 K (Kf = 4.9). Find the degree of association.
- Observed M = 4.9 × 2.0 × 1000/(1.62 × 25) = 242 g mol⁻¹.
- i = 122/242 = 0.504.
- Dimerisation, n = 2: α = (1 − 0.504)/(1 − 0.5).
Answer: α ≈ 0.99 — almost all of it exists as hydrogen-bonded dimers.
Check yourself
- C3. Which 0.1 m solution freezes lowest: glucose, NaCl or CaCl₂?
- C4. Why does acetic acid show a molar mass near 120 in benzene?
Answers at the end.
Exam trap gallery
The mistakes that cost the most marks in this chapter.
| # | What students write | What is actually correct | Why |
|---|---|---|---|
| 1 | Use molarity in ΔTf = Kf × m. | Use molality (mol per kg of solvent). | Molality doesn’t change with temperature; the K values are defined for it. |
| 2 | 0.1 m NaCl and 0.1 m glucose have the same ΔTf. | NaCl gives nearly twice the ΔTf (i ≈ 2). | Colligative properties count particles, not formula units. |
| 3 | Higher K(H) means the gas is more soluble. | Higher K(H) means less soluble: x = p/K(H). | K(H) is in the denominator. |
| 4 | ΔTf is added to 0 °C. | The solution freezes below 0 °C: T = 0 − ΔTf. | ΔTf is a depression. |
| 5 | An azeotrope is a compound. | It’s a mixture; its composition changes with pressure. | It only boils at constant composition at one pressure. |
| 6 | Chloroform + acetone shows positive deviation. | Negative deviation (maximum-boiling azeotrope). | H bonding between them makes A–B attraction stronger. |
| 7 | i is always a whole number. | i is usually fractional (e.g. 1.87 for NaCl). | Dissociation is incomplete, especially at higher concentration. |
| 8 | In π = CRT, use R = 8.314 with C in mol L⁻¹. | Use R = 0.083 L bar K⁻¹ mol⁻¹ (or 0.0821 L atm) for π in bar (atm). | Units must match: 8.314 needs SI units throughout. |
Memory hook: “Count the crowd, not the faces” — colligative = number of particles. · “Molal for heat, molar for pressure” — ΔTb/ΔTf use m; π uses C. · “Big K(H), shy gas” — high Henry’s constant, low solubility. · “Plus pushes out, minus holds in” — positive deviation, higher VP; negative deviation, lower VP.
Practice MCQs
Levels: Easy checks recall, Moderate needs one reasoning step, Tough combines two ideas.
Q01. Which concentration term depends on temperature? (Easy)
- (a) Molality
- (b) Mole fraction
- (c) Molarity
- (d) Mass percent
Q02. The unit of the molal elevation constant Kb is: (Easy)
- (a) K mol kg⁻¹
- (b) K kg mol⁻¹
- (c) mol kg K⁻¹
- (d) K⁻¹ kg mol⁻¹
Q03. Which method is best for finding the molar mass of a protein? (Easy)
- (a) Elevation of boiling point
- (b) Depression of freezing point
- (c) Osmotic pressure
- (d) Relative lowering of VP
Q04. A solution of chloroform and acetone shows: (Easy)
- (a) positive deviation
- (b) negative deviation
- (c) ideal behaviour
- (d) no deviation at any composition
Q05. Which 0.1 m aqueous solution has the highest boiling point? (Easy)
- (a) Glucose
- (b) KCl
- (c) K₂SO₄
- (d) Urea
Q06. The van’t Hoff factor of acetic acid fully dimerised in benzene is: (Moderate)
- (a) 2
- (b) 1
- (c) 0.5
- (d) 0.25
Q07. The mole fraction of solute in a 1 molal aqueous solution is: (Moderate)
- (a) 0.0177
- (b) 0.0180
- (c) 0.1770
- (d) 1.0000
Q08. A solution showing positive deviation from Raoult’s law forms: (Moderate)
- (a) a maximum-boiling azeotrope
- (b) a minimum-boiling azeotrope
- (c) no azeotrope
- (d) a solid solution
Q09. 1.0 g urea (M = 60) in 100 g water: ΔTf is (Kf = 1.86): (Moderate)
- (a) 0.031 K
- (b) 0.31 K
- (c) 3.1 K
- (d) 0.186 K
Q10. A 0.1 m solution of a salt AB₂ shows i = 2.6. The degree of dissociation is: (Moderate)
- (a) 0.6
- (b) 0.8
- (c) 0.9
- (d) 0.53
Q11. Red blood cells placed in 2% NaCl solution will: (Moderate)
- (a) swell
- (b) burst
- (c) shrink
- (d) remain unchanged
Q12. For an ideal solution, which is NOT true? (Moderate)
- (a) ΔH(mix) = 0
- (b) ΔV(mix) = 0
- (c) It obeys Raoult’s law
- (d) ΔS(mix) = 0
Q13. In a mixture of heptane (p° = 105.2 kPa) and octane (p° = 46.8 kPa) with x(heptane) = 0.459, the mole fraction of heptane in the vapour is about: (Tough)
- (a) 0.46
- (b) 0.54
- (c) 0.66
- (d) 0.34
Q14. 2 g benzoic acid in 25 g benzene gives ΔTf = 1.62 K (Kf = 4.9). The percentage association is about: (Tough)
- (a) 50%
- (b) 75%
- (c) 99%
- (d) 25%
Q15. 10 g of a solute in 90 g water lowers the vapour pressure by 2%. Its molar mass is about: (Tough)
- (a) 49
- (b) 98
- (c) 196
- (d) 9.8
Answer key with explanations
- (c) Molarity — Molarity uses solution volume, which expands on heating.
- (b) K kg mol⁻¹ — ΔTb = Kb × m, so Kb = K/(mol kg⁻¹) = K kg mol⁻¹.
- (c) Osmotic pressure — π is large enough to measure for dilute solutions, at room temperature.
- (b) negative deviation — Hydrogen bonding between the two makes A–B attraction stronger.
- (c) K₂SO₄ — K₂SO₄ gives 3 ions (i ≈ 3).
- (c) 0.5 — Two molecules become one particle: i = 1/2.
- (a) 0.0177 — 1/(1 + 55.5) = 0.0177.
- (b) a minimum-boiling azeotrope — Higher vapour pressure → lower boiling point at the azeotropic composition.
- (b) 0.31 K — m = 0.167; ΔTf = 1.86 × 0.167 = 0.31 K.
- (b) 0.8 — AB₂ → A²⁺ + 2B⁻, n = 3: α = (2.6 − 1)/(3 − 1) = 0.8.
- (c) shrink — 2% NaCl is hypertonic compared with 0.9%; water leaves the cells.
- (d) ΔS(mix) = 0 — Mixing always increases entropy: ΔS(mix) > 0.
- (c) 0.66 — p(hep) = 48.3 kPa, p(total) = 73.6 kPa; y = 48.3/73.6 = 0.66.
- (c) 99% — Observed M = 242; i = 0.504; α = (1 − 0.504)/0.5 = 0.99.
- (b) 98 — n₂/(n₁ + n₂) = 0.02 with n₁ = 5 → n₂ = 0.102 → M = 98.
Advanced problems
TGT/PGT level: each needs two ideas from the chapter.
Solved example 10 — Freezing point from osmotic pressure
A dilute aqueous solution has π = 2.49 bar at 300 K. Estimate its freezing point (Kf = 1.86; assume molality ≈ molarity).
- C = π/RT = 2.49/(0.083 × 300) = 0.100 mol L⁻¹.
- For a dilute aqueous solution, m ≈ C = 0.100 mol kg⁻¹.
- ΔTf = 1.86 × 0.100 = 0.186 K.
Answer: About −0.19 °C. Any one colligative property predicts the others.
Solved example 11 — Choosing between two salts
Which lowers the freezing point of water more per gram: NaCl (M = 58.5) or CaCl₂ (M = 111)? Assume complete dissociation.
- Particles per gram, NaCl: 2/58.5 = 0.0342 mol.
- Particles per gram, CaCl₂: 3/111 = 0.0270 mol.
Answer: NaCl gives more particles per gram, so it lowers the freezing point more per gram (CaCl₂ is still used on roads because it works at lower temperatures and releases heat as it dissolves).
Check-yourself answers
- C1. Molarity.
- C2. Benzene is more volatile (higher p°), and y = p(A)/p(total) favours the more volatile component.
- C3. CaCl₂ (i ≈ 3).
- C4. It dimerises through hydrogen bonds in benzene, so the number of particles halves and the apparent molar mass doubles.
One-page revision
Read this page the night before the exam.
Concentration
- Molarity changes with T; molality, x, mass % don’t
- 1 molal aqueous → x(solute) = 0.0177
- Use molality for ΔTb, ΔTf; molarity for π
Henry and Raoult
- p = K(H) x; big K(H) = less soluble
- p(A) = x(A) p°(A)
- Vapour richer in the more volatile liquid
- (p° − p)/p° = x(solute)
Deviations
- Positive: weaker A–B, higher VP, ΔH > 0, min-boiling azeotrope
- Negative: stronger A–B, lower VP, ΔH < 0, max-boiling azeotrope
- CHCl₃ + acetone negative; ethanol + acetone positive
Colligative
- ΔTb = i Kb m; ΔTf = i Kf m; π = iCRT
- Water: Kb 0.52, Kf 1.86
- M = K w₂ 1000/(ΔT w₁)
- π is best for proteins and polymers
van’t Hoff factor
- Dissociation: α = (i − 1)/(n − 1)
- Association: α = (1 − i)/(1 − 1/n)
- Dimer in benzene → i ≈ 0.5
- 0.1 m CaCl₂ > NaCl > glucose for ΔTf
Osmosis
- Solvent moves dilute → concentrated
- Isotonic 0.9% NaCl; hypertonic shrinks cells
- Reverse osmosis desalinates water
- Memory: count the crowd, not the faces