Why colligative properties go abnormal
The four colligative properties — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure — all depend only on the number of solute particles in solution, not on their chemical nature. The standard formulae assume that one mole of dissolved solute contributes exactly one mole of independent particles. Whenever a solute breaks up into ions, or two or more molecules join into a single aggregate, that assumption fails, and the measured property no longer matches the value calculated from the weighed mass.
Consider potassium chloride. One mole of KCl (74.5 g) dissolved in water releases one mole of $\ce{K+}$ and one mole of $\ce{Cl-}$, giving two moles of particles. If we ignore interionic attractions, one mole of KCl in one kg of water raises the boiling point by $2 \times 0.52 = 1.04\ \text{K}$ instead of the $0.52\ \text{K}$ expected for a single particle. Working backwards from the larger observed elevation, one would conclude that two moles of particles weigh 74.5 g, so the apparent molar mass of KCl is only 37.25 g — half the true value.
Dissociation lowers the apparent molar mass; association raises it
When a solute dissociates into ions, the particle count rises, the colligative property is larger than expected, and the experimentally determined molar mass is always lower than the true value. When a solute associates into dimers or larger units, the particle count falls, the property is smaller than expected, and the determined molar mass is higher than the true value.
Both lower-than-normal and higher-than-normal results are called abnormal molar mass.
The mirror case is association. Ethanoic (acetic) acid dimerises in benzene through hydrogen bonding — a behaviour typical of solvents of low dielectric constant — so the number of independent particles is reduced. If every molecule paired up, the depression of freezing point or elevation of boiling point would be exactly half the normal value, and the molar mass calculated from that property would be twice the expected figure. The association equilibrium is written as:
$\ce{2 CH3COOH <=> (CH3COOH)2}$
Defining the van't Hoff factor
In 1880 van't Hoff introduced a single correction factor, i, to account for the extent of dissociation or association. It can be written in three equivalent forms, all of which appear in NCERT:
$$ i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} = \frac{\text{Total moles of particles after assoc./dissoc.}}{\text{Moles of particles before assoc./dissoc.}} $$
Here the abnormal molar mass is the value experimentally determined from a colligative property, while the calculated property is the one obtained by assuming the non-volatile solute is neither associated nor dissociated. The three expressions are consistent because the colligative property is inversely proportional to molar mass and directly proportional to particle count.
| Behaviour of solute | Particle count | Value of i | Apparent molar mass |
|---|---|---|---|
| No dissociation / association | Unchanged | i = 1 | Equal to true value |
| Dissociation (e.g. NaCl, KCl) | Increases | i > 1 | Lower than true value |
| Association (e.g. acetic acid in benzene) | Decreases | i < 1 | Higher than true value |
As reference points NCERT quotes i for aqueous KCl as close to 2, and i for ethanoic acid in benzene as nearly 0.5. These two numbers anchor the entire concept: a number above 1 signals dissociation, a number below 1 signals association, and exactly 1 means the solute stays intact as discrete molecules.
Modified colligative formulae
Because every colligative property scales with the number of particles, inserting i as a multiplier corrects each formula. The four NCERT expressions become:
| Colligative property | Normal formula | Modified formula (with i) |
|---|---|---|
| Relative lowering of vapour pressure | $\dfrac{p^\circ - p}{p^\circ} = \dfrac{n_2}{n_1}$ | $\dfrac{p^\circ - p}{p^\circ} = i\,\dfrac{n_2}{n_1}$ |
| Elevation of boiling point | $\Delta T_b = K_b\, m$ | $\Delta T_b = i\,K_b\, m$ |
| Depression of freezing point | $\Delta T_f = K_f\, m$ | $\Delta T_f = i\,K_f\, m$ |
| Osmotic pressure | $\Pi = \dfrac{n_2 R T}{V} = C R T$ | $\Pi = i\,\dfrac{n_2 R T}{V} = i\,C R T$ |
The single takeaway is that every colligative property is directly proportional to i. Two solutions of the same molal (or molar) concentration will show colligative effects in the ratio of their i values, which is the structural basis of almost every ranking question.
The factor i only rescales a property you already understand. If $\Pi = iCRT$ still feels unfamiliar, revisit Osmotic Pressure before attempting electrolyte problems.
Degree of dissociation and association
The van't Hoff factor connects directly to how completely a solute splits or pairs. The degree of dissociation, $\alpha$, is the fraction of dissolved formula units that break into ions; the degree of association is the fraction that combine into aggregates. Both are obtained from the particle-count form of i.
Dissociation
Take a solute that dissociates into $n$ particles, for example $\ce{CH3COOH <=> CH3COO^- + H+}$ (here $n = 2$). Starting from one mole, if $\alpha$ is the degree of dissociation, then at equilibrium there are $(1-\alpha)$ mol undissociated solute and $n\alpha$ mol of ions, giving a total of $1 - \alpha + n\alpha = 1 + (n-1)\alpha$ moles of particles. Since this total equals i:
$$ i = 1 + (n-1)\alpha \qquad\Longrightarrow\qquad \alpha = \frac{i - 1}{n - 1} $$
Association
Now take a solute where $n$ molecules associate into one aggregate, for example $\ce{2 CH3COOH <=> (CH3COOH)2}$ (here $n = 2$). Starting from one mole, if $\alpha$ is the degree of association, then $\alpha$ mol associate into $\alpha/n$ mol of aggregate while $(1-\alpha)$ mol remain free, giving a total of $1 - \alpha + \alpha/n$ moles of particles, which equals i:
$$ i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha\left(1 - \frac{1}{n}\right) \qquad\Longrightarrow\qquad \alpha = \frac{1 - i}{\,1 - \tfrac{1}{n}\,} $$
Use the right formula for the right direction
For dissociation, $\alpha = \dfrac{i-1}{n-1}$ where $n$ is the number of ions produced. For association, $\alpha = \dfrac{1-i}{1-1/n}$ where $n$ is the number of molecules in one aggregate. A common slip is using the dissociation formula for a dimerising acid, which gives a negative answer because $i < 1$ there.
Master table of i values
The table below collects the standard solutes referenced in NCERT and the NIOS supplement, with their behaviour and the limiting value of i for complete dissociation or full association. The starred values for strong electrolytes are the actual measured i at moderate concentration, which lie below the integer limit because dissociation is incomplete and interionic attraction reduces the effective particle count.
| Solute | Process in solution | Particles produced | i (limiting) |
|---|---|---|---|
| Glucose, urea, sucrose | None (non-electrolyte) | 1 | 1 |
| $\ce{NaCl}$ | Dissociation | $\ce{Na+ + Cl-}$ | 2 |
| $\ce{KCl}$ | Dissociation | $\ce{K+ + Cl-}$ | 2 |
| $\ce{MgSO4}$ | Dissociation | $\ce{Mg^2+ + SO4^2-}$ | 2 |
| $\ce{CaCl2}$ | Dissociation | $\ce{Ca^2+ + 2Cl-}$ | 3 |
| $\ce{K2SO4}$ | Dissociation | $\ce{2K+ + SO4^2-}$ | 3 |
| $\ce{AlCl3}, \ce{FeCl3}$ | Dissociation | 1 cation + 3 anions | 4 |
| Ethanoic acid in benzene | Association (dimer) | $\ce{(CH3COOH)2}$ | ≈ 0.5 |
| Benzoic acid in benzene | Association (dimer) | $\ce{(C6H5COOH)2}$ | ≈ 0.5 |
NCERT Table 1.4 makes the dilution trend explicit: for NaCl, i rises from 1.87 at 0.1 m to 1.97 at 0.001 m; for KCl it goes 1.85 → 1.98; for $\ce{MgSO4}$ it climbs steeply 1.21 → 1.82; and for $\ce{K2SO4}$ it moves 2.32 → 2.84 toward its limit of 3. The factor approaches its ideal integer value as the solution becomes more dilute, because interionic attraction weakens.
Worked examples
0.6 mL of acetic acid ($\ce{CH3COOH}$, density 1.06 g mL⁻¹) is dissolved in 1 litre of water. The observed depression in freezing point is 0.0205 °C. Take $K_f = 1.86\ \text{K kg mol}^{-1}$. Find the van't Hoff factor and the degree of dissociation.
Moles of acetic acid $= \dfrac{0.6 \times 1.06}{60} = 0.0106\ \text{mol}$, so molality $\approx 0.0106\ \text{mol kg}^{-1}$.
Calculated depression: $\Delta T_f = K_f\, m = 1.86 \times 0.0106 = 0.0197\ \text{K}$.
van't Hoff factor: $i = \dfrac{\text{observed}}{\text{calculated}} = \dfrac{0.0205}{0.0197} = 1.041$.
Degree of dissociation: acetic acid splits into two ions, so $n = 2$ and $\alpha = \dfrac{i - 1}{n - 1} = 1.041 - 1.000 = 0.041$. About 4.1% of the acid is dissociated, consistent with its weak-acid character.
2 g of benzoic acid ($\ce{C6H5COOH}$, molar mass 122 g mol⁻¹) dissolved in 25 g of benzene shows a freezing-point depression of 1.62 K. The molal depression constant for benzene is $K_f = 4.9\ \text{K kg mol}^{-1}$. Find the percentage association if the acid forms a dimer.
Experimental (abnormal) molar mass: $M_2 = \dfrac{K_f \times w_2 \times 1000}{w_1 \times \Delta T_f} = \dfrac{4.9 \times 2 \times 1000}{25 \times 1.62} = 241.98\ \text{g mol}^{-1}$.
van't Hoff factor: $i = \dfrac{\text{normal molar mass}}{\text{abnormal molar mass}} = \dfrac{122}{241.98} = 0.504$.
Degree of association: for a dimer $n = 2$, so $\alpha = \dfrac{1 - i}{1 - 1/2} = \dfrac{1 - 0.504}{0.5} = 0.992$. Hence about 99.2% of the benzoic acid is associated as dimer in benzene.
Ranking equimolar solutes
The most frequently tested skill is ordering colligative effects of equimolar solutes. Because every colligative property is proportional to i, the ranking reduces to ranking the i values, provided concentrations are equal. The product $i \times m$ (or $i \times C$) decides the magnitude of any colligative property.
Arrange equimolal aqueous solutions of glucose, $\ce{NaCl}$ and $\ce{CaCl2}$ in increasing order of (a) freezing point and (b) boiling point.
i values: glucose $i \approx 1$; $\ce{NaCl -> Na+ + Cl-}$, $i \approx 2$; $\ce{CaCl2 -> Ca^2+ + 2Cl-}$, $i \approx 3$.
Colligative magnitude $\Delta T \propto i$, so the depression and elevation both follow $\ce{CaCl2} > \ce{NaCl} > \text{glucose}$.
(a) Freezing point: larger depression means lower freezing point, so freezing point increases as $\ce{CaCl2} < \ce{NaCl} < \text{glucose}$.
(b) Boiling point: larger elevation means higher boiling point, so boiling point increases as $\text{glucose} < \ce{NaCl} < \ce{CaCl2}$.
i decides everything, but watch which direction the property moves
Larger i always means a larger colligative effect, hence a higher osmotic pressure, larger boiling-point elevation and larger freezing-point depression. But a larger depression means a lower freezing point, so the freezing-point ranking is the reverse of the boiling-point ranking. For osmotic pressure and boiling point, follow $i$ directly: $\ce{CaCl2} > \ce{NaCl} > \text{glucose}$.
Rule of thumb: rank by $i \times m$. Then flip the order only when the question asks for freezing point itself rather than its depression.
Van't Hoff factor in one screen
- $i = \dfrac{\text{normal molar mass}}{\text{abnormal molar mass}} = \dfrac{\text{observed colligative property}}{\text{calculated colligative property}} = \dfrac{\text{particles after}}{\text{particles before}}$.
- $i > 1$ for dissociation (more particles, lower apparent molar mass); $i < 1$ for association (fewer particles, higher apparent molar mass); $i = 1$ for non-electrolytes.
- Every modified formula carries $i$: $\Delta T_b = iK_b m$, $\Delta T_f = iK_f m$, $\Pi = iCRT$, and $\frac{p^\circ-p}{p^\circ}=i\frac{n_2}{n_1}$.
- Degree of dissociation $\alpha = \dfrac{i-1}{n-1}$; degree of association $\alpha = \dfrac{1-i}{1-1/n}$.
- Limiting $i$: glucose/urea/sucrose = 1; NaCl, KCl, $\ce{MgSO4}$ = 2; $\ce{CaCl2}$, $\ce{K2SO4}$ = 3; dimerising acids in benzene ≈ 0.5.
- Rank colligative effects by $i\times m$: $\ce{CaCl2} > \ce{NaCl} > \text{glucose}$ at equal concentration.