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This MCQ module is based on: Colligative Vant Hoff
TOPIC 3 OF 23
Colligative Vant Hoff
🎓 Class 12
Chemistry
CBSE
Theory
Ch 1 – Solutions
⏱ ~14 min
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Colligative Vant Hoff
What Makes a Property "Colligative"?
Dissolve a pinch of sugar in a cup of hot tea: the boiling point of the tea actually rises a tiny bit, and when the cup cools, the tea freezes at a slightly lower temperature than pure water. Remarkably, it does not matter whether the solute is sugar or urea or glycine — what matters is only how many particles are dissolved. Properties that depend on the number of solute particles, and not on their nature, are called colligative properties.
There are four of them, and each one gives an independent route to the molar mass of the solute:
- Relative lowering of vapour pressure
- Elevation of boiling point
- Depression of freezing point
- Osmotic pressure
1.6 Colligative Properties and Determination of Molar Mass
1.6.1 Relative Lowering of Vapour Pressure
Put a non-volatile solute (B) into a volatile solvent (A). Some solute particles occupy the surface, blocking solvent molecules from escaping. The solvent's vapour pressure drops from pA° to pA. By Raoult's law:
\(p_{A}=p_{A}^{\circ}\,x_{A}\quad\Rightarrow\quad \dfrac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}} = 1-x_{A} = x_{B}\)
The left side is the relative lowering of vapour pressure and equals the mole fraction of the solute. For dilute solutions nB << nA, giving:
\(\dfrac{\Delta p}{p_{A}^{\circ}} \;\approx\; \dfrac{n_{B}}{n_{A}} \;=\; \dfrac{w_{B}/M_{B}}{w_{A}/M_{A}}\)
from which the molar mass MB of the solute can be extracted if we know the masses (wA, wB) and the molar mass of the solvent.
1.6.2 Elevation of Boiling Point
A liquid boils when its vapour pressure equals the external pressure. Since vapour pressure is lowered, a solution must be heated higher than the pure solvent to reach 1 atm. For dilute solutions:
\(\Delta T_{b} \;=\; K_{b}\,m\)
where m is molality and Kb is the molal elevation constant (ebullioscopic constant, units K kg mol⁻¹). For water Kb = 0.52 K kg mol⁻¹; so a 1 m aqueous solution of a non-volatile non-electrolyte boils at 100.52 °C.
1.6.3 Depression of Freezing Point
At the freezing point, solid solvent and liquid solvent coexist at the same vapour pressure. In solution, the liquid's vapour pressure is lowered, so the new freezing point must be below that of pure solvent.
\(\Delta T_{f} \;=\; K_{f}\,m\)
where Kf is the molal depression constant (cryoscopic constant). For water Kf = 1.86 K kg mol⁻¹. Practical uses:
- Antifreeze in car radiators: 30 % ethylene glycol lowers freezing point to about −18 °C.
- De-icing roads: NaCl or CaCl₂ depresses the freezing point of water on roads from 0 °C to well below.
- Ice-cream making: the ice–salt bath outside the churn can reach −20 °C, fast enough to freeze the cream inside.
1.6.4 Osmosis and Osmotic Pressure
If a solution and its pure solvent are separated by a semi-permeable membrane (which lets only solvent through), solvent molecules flow spontaneously from the pure side into the solution, diluting it. This flow is osmosis. The extra hydrostatic pressure required on the solution side to stop the flow is called the osmotic pressure, π.
Van't Hoff equation (for dilute solutions):
\(\pi \;=\; \dfrac{n_{B}}{V}\,RT \;=\; C\,RT\)
C = molar concentration of the solution, R = 0.0821 L atm mol⁻¹ K⁻¹ (or 8.314 J mol⁻¹ K⁻¹), T in kelvin.Isotonic, hypertonic and hypotonic
- Isotonic solutions have the same π (e.g., 0.9 % w/v NaCl — "normal saline" — is isotonic with human blood).
- Hypertonic: higher π than the cell inside. Water flows out, the cell shrinks (plasmolysis). Seawater is hypertonic to red blood cells.
- Hypotonic: lower π. Water rushes into the cell, which swells and may burst (haemolysis).
Reverse Osmosis
If an external pressure greater than π is applied on the solution side, solvent actually flows from the solution to the pure side — this is reverse osmosis. It is the principle behind most modern desalination plants: sea water pushed against a special polyamide membrane at ~30–40 atm yields fresh drinking water on the other side.
Worked Examples — Colligative Properties
Example 1.14 — Relative lowering of VP from urea in water
18.0 g of urea (M = 60) is dissolved in 180 g of water. Calculate the relative lowering of vapour pressure at 298 K. (The pure water vapour pressure need not be used numerically.)
\(n_{urea}=\dfrac{18}{60}=0.30\,\text{mol};\quad n_{water}=\dfrac{180}{18}=10.0\,\text{mol}\)
\(\dfrac{\Delta p}{p^{\circ}} = x_{urea} = \dfrac{0.30}{0.30+10.0}=0.02913\)
Answer: Δp/p° = 0.0291 or ≈ 2.91 %.
Example 1.15 — Boiling-point elevation of a sugar solution
Calculate the boiling point of a solution containing 18 g of glucose (M = 180) in 100 g of water. Kb (water) = 0.52 K kg mol⁻¹.
\(m = \dfrac{18/180}{0.100}=1.00\,\text{mol kg}^{-1}\)
\(\Delta T_{b}=0.52\times1.00=0.52\,\text{K}\)
Boiling point = 373.15 + 0.52 = 373.67 K (100.52 °C).
Example 1.16 — Antifreeze calculation
What mass of ethylene glycol (M = 62) must be added to 4 kg of water to lower its freezing point to −6 °C? Kf(water) = 1.86 K kg mol⁻¹.
\(\Delta T_{f}=6\,\text{K}\;\Rightarrow\;m=\dfrac{\Delta T_{f}}{K_{f}}=\dfrac{6}{1.86}=3.226\,\text{mol kg}^{-1}\)
\(n_{glycol}=3.226\times 4=12.90\,\text{mol};\; w=12.90\times 62=800\,\text{g}\)
Answer: about 800 g of ethylene glycol are required.
Example 1.17 — Molar mass of a protein from osmotic pressure
200 cm³ of an aqueous solution containing 1.26 g of a protein exerts π = 2.57 × 10⁻³ bar at 300 K. Determine the molar mass of the protein.
\(\pi V = nRT \;\Rightarrow\; n=\dfrac{\pi V}{RT}\)
Using R = 0.0831 L bar mol⁻¹ K⁻¹ and V = 0.200 L:
\(n=\dfrac{2.57\times10^{-3}\times0.200}{0.0831\times300}=2.062\times10^{-5}\,\text{mol}\)
\(M=\dfrac{w}{n}=\dfrac{1.26}{2.062\times10^{-5}}\approx 6.11\times10^{4}\,\text{g mol}^{-1}\)
Answer: ≈ 61 100 g mol⁻¹ — typical of a medium-sized protein. Osmotic pressure is the method of choice for macromolecules because small Δp, ΔTb, ΔTf would be too tiny to measure.
1.7 Abnormal Molar Masses and the van't Hoff Factor
Our colligative equations assume one dissolved particle per solute formula unit. But when NaCl dissolves it dissociates into Na⁺ + Cl⁻ — two particles. When benzoic acid dissolves in benzene it associates into dimers held by hydrogen bonds — half a particle on average. The observed colligative effect then differs from the calculated value.
Van't Hoff factor (i):
\(i=\dfrac{\text{observed colligative property}}{\text{calculated value assuming no dissociation/association}}=\dfrac{\text{actual number of particles}}{\text{number of formula units dissolved}}\)
Modified colligative equations:
\(\dfrac{\Delta p}{p^{\circ}}=i\,x_{B},\quad \Delta T_{b}=i\,K_{b}\,m,\quad \Delta T_{f}=i\,K_{f}\,m,\quad \pi=i\,C\,R\,T\)
Dissociation (ionic solutes)
NaCl → Na⁺ + Cl⁻, so i ≈ 2. K₂SO₄ → 2K⁺ + SO₄²⁻, i ≈ 3. If the dissociation is incomplete (degree of dissociation α for an n-ion electrolyte):
\(i = 1+(n-1)\alpha \quad\text{(for dissociation)}\)
Association
Benzoic acid in benzene: 2 C₆H₅COOH ⇌ (C₆H₅COOH)₂. Two molecules merge into one dimer, giving i ≈ 0.5. If the degree of association (fraction of monomers that pair up to form an n-mer) is α:
\(i = 1-\alpha\left(1-\dfrac{1}{n}\right) \quad\text{(for association)}\)
| Solute | Solvent | i (typical) | Why |
|---|---|---|---|
| Glucose, urea, sucrose | water | 1.00 | Non-electrolyte; no change in particle count |
| NaCl | water | ≈ 2 (1.8–2.0) | Dissociates to Na⁺ + Cl⁻ |
| MgCl₂, K₂SO₄ | water | ≈ 3 | Dissociates to three ions |
| Fe₂(SO₄)₃ | water | ≈ 5 | Dissociates to 2 Fe³⁺ + 3 SO₄²⁻ |
| Benzoic acid | benzene | ≈ 0.5 | Dimerises via H-bonds |
| Acetic acid | benzene | ≈ 0.5 | Dimerises |
Worked Examples — van't Hoff Factor
Example 1.18 — van't Hoff factor for NaCl
A 0.100 m NaCl solution in water shows ΔTf = 0.348 K. Find i. (Kf = 1.86.)
\(\Delta T_{f,\,calc}=K_{f}\,m=1.86\times0.100=0.186\,\text{K}\)
\(i=\dfrac{0.348}{0.186}=1.87\)
Interpretation: NaCl is not 100 % dissociated at this concentration — ion-pairing reduces i slightly below 2. Using i = 1 + α gives α ≈ 0.87 (87 % dissociation).
Example 1.19 — Degree of dissociation of CH₃COOH
0.0106 mol of CH₃COOH in 1 kg water lowers the freezing point by 0.0205 K. Find the degree of dissociation α.
\(\Delta T_{f,\,calc}=1.86\times0.0106=0.01972\,\text{K}; \quad i=\dfrac{0.0205}{0.01972}=1.040\)
For CH₃COOH → CH₃COO⁻ + H⁺, n = 2, so i = 1 + α:
\(\alpha = i-1 = 0.040 \;\; (4.0\,\%)\)
Consistent with acetic acid being a weak acid.
Example 1.20 — Association of benzoic acid in benzene
A solution of 0.400 g benzoic acid (M = 122) in 30 g of benzene depresses the freezing point by 0.26 K. Kf(benzene) = 5.12. Find the degree of association α assuming dimerisation.
\(m=\dfrac{0.400/122}{0.030}=0.1093\,\text{mol kg}^{-1}\)
\(\Delta T_{f,\,calc}=5.12\times0.1093=0.5596\,\text{K};\;i=\dfrac{0.26}{0.5596}=0.4646\)
For 2 A → A₂, use i = 1 − α(1 − 1/2) = 1 − α/2:
\(0.4646=1-\dfrac{\alpha}{2}\;\Rightarrow\;\alpha = 1.071 \approx 1.0\)
Answer: essentially all benzoic acid molecules are dimerised in benzene (α ≈ 1).
Example 1.21 — Combined problem
A 5 % (w/v) aqueous solution of sucrose (M = 342) is isotonic with a 0.877 % (w/v) aqueous solution of an unknown electrolyte X which dissociates as X → 2 X⁺ + X²⁻. Find the molar mass of X.
Isotonic → equal π → equal i·C.
\(C_{sucrose}=\dfrac{50}{342}=0.1462\,\text{mol L}^{-1};\;(i_{sucrose}=1)\)
For X, i = 3 (three ions). Let MX be the molar mass; CX = 8.77/MX.
\(3\cdot\dfrac{8.77}{M_{X}}=0.1462\;\Rightarrow\;M_{X}=\dfrac{3\times8.77}{0.1462}=179.9\,\text{g mol}^{-1}\)
Answer: MX ≈ 180 g mol⁻¹.
Activity 1.3 — Salt on Ice
L3 Apply
Predict: Place an ice cube on a plate and sprinkle common salt over it. Will the ice (a) melt faster, (b) melt slower, or (c) no change? Why?
- Take two ice cubes of similar size on two plates.
- Sprinkle a teaspoon of salt on one ice cube; leave the other as a control.
- Time how long each cube takes to fully melt.
- Feel the liquid that forms under the salted cube — is it colder than plain melt-water?
The salted cube melts faster, and the resulting brine is colder than 0 °C — often around −5 to −8 °C. Salt depresses the freezing point of water; since room heat still flows in at the same rate, melting proceeds. The same principle is behind ice-cream making and road de-icing. Magnesium chloride (CaCl₂) works even better because each formula unit releases 3 particles, so i ≈ 3.
Interactive: Colligative Property Calculator L3 Apply
Property:
Mass of solute (g):
M of solute (g/mol):
Mass of solvent (g):
i (van't Hoff):
K (Kb or Kf, K kg/mol):
V (L) for π:
T (K) for π:
Enter data and press Compute.
Competency-Based Questions
A pharmacist prepares an IV glucose drip (5 % w/v glucose in water, Mglucose = 180) and a 0.9 % w/v normal saline. A separate student dissolves 0.10 mol KCl in 1 kg water and measures a freezing-point depression of 0.35 K.
Q1. L1 Remember Which one is NOT a colligative property?
Answer: C. The vapour pressure of pure solvent is a property of the solvent alone; colligative properties describe how solutes change solvent properties.
Q2. L3 Apply Compute the osmotic pressure of the 5 % glucose drip at 310 K. (3 marks)
C = 50/180 = 0.278 mol L⁻¹; i = 1. π = CRT = 0.278 × 0.0821 × 310 = 7.07 atm. This is close to the osmotic pressure of blood (~7.6 atm), explaining why it is safe for intravenous use.
Q3. L3 Apply For the KCl experiment, find the van't Hoff factor and the degree of dissociation. (Kf = 1.86.) (3 marks)
ΔTf,calc = 1.86 × 0.10 = 0.186 K. i = 0.35/0.186 = 1.88. For KCl, n = 2, so i = 1 + α → α = 0.88 (88 % dissociation).
Q4. L4 Analyse A red blood cell placed in distilled water bursts. Placed in 10 % NaCl solution it shrivels. Explain both observations using the terms hypotonic, hypertonic and isotonic. (3 marks)
Distilled water is hypotonic to the cell: π outside < π inside, so water rushes in, swelling the cell until it bursts (haemolysis). 10 % NaCl is strongly hypertonic: π outside ≫ π inside, so water leaves the cell, shrivelling it (plasmolysis). Normal saline (0.9 % NaCl) is isotonic with blood and leaves cells unchanged — which is why IV drips use it.
Q5. L5 Evaluate For determining the molar mass of a 50 000 g mol⁻¹ protein, would you choose ΔTb, ΔTf or π? Justify. (3 marks)
Osmotic pressure. For a 1 g/L aqueous solution of such a protein, m = 2 × 10⁻⁵ mol/kg so ΔTb = 10⁻⁵ K and ΔTf = 4 × 10⁻⁵ K — unmeasurably small. Yet π = CRT = 2 × 10⁻⁵ × 0.0821 × 300 ≈ 5 × 10⁻⁴ atm = 5 mm of water column — easily measured. Osmotic methods are therefore the standard for macromolecules.
Assertion-Reason Questions
Assertion (A): 0.1 m NaCl lowers the freezing point of water roughly twice as much as 0.1 m glucose.
Reason (R): NaCl dissociates in water into Na⁺ and Cl⁻, roughly doubling the number of particles per formula unit.
Answer: A. Colligative properties depend on particle count. Each mole of NaCl gives ~2 moles of ions; glucose gives 1.
Assertion (A): Benzoic acid shows a van't Hoff factor of about 0.5 in benzene.
Reason (R): Benzoic acid molecules associate through hydrogen bonding to form dimers in non-polar solvents.
Answer: A. Two monomers merge to one dimer → number of particles halves → i ≈ 0.5.
Assertion (A): Reverse osmosis is used in desalination plants to obtain drinkable water from sea water.
Reason (R): Applying a pressure greater than the osmotic pressure of sea water pushes pure water through a semi-permeable membrane from the solution to the solvent side.
Answer: A. Reverse osmosis is precisely the reversal of natural solvent flow by imposing P > π.
Frequently Asked Questions - Colligative Vant Hoff
What is the main concept covered in Colligative Vant Hoff?
In NCERT Class 12 Chemistry Chapter 1 (Solutions), "Colligative Vant Hoff" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Colligative Vant Hoff useful in real-life or applied chemistry?
Real-life applications of "Colligative Vant Hoff" from NCERT Class 12 Chemistry Chapter 1 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Colligative Vant Hoff?
Key reactions in "Colligative Vant Hoff" (NCERT Class 12 Chemistry Chapter 1 Solutions) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 1?
NCERT Class 12 Chemistry Chapter 1 (Solutions) is structured so each part builds chemical understanding sequentially. "Colligative Vant Hoff" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Colligative Vant Hoff?
CBSE board questions from "Colligative Vant Hoff" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Colligative Vant Hoff" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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