JEE Mains Chemistry · Solutions

Van't Hoff Factor and Abnormal Molar Mass

The van 't Hoff factor i is the number of particles a formula unit really gives in solution; it multiplies every colligative effect, exceeds 1 for dissociation and falls below 1 for association.

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Why this matters

Twenty-eight PYQs, the largest page, eighteen of them numeric, and two from 2026. Seven rank solutions by boiling or freezing point, which is a count of particles; sixteen find i, a degree of dissociation or an acid's Ka from a measured shift; five treat a solute that associates, such as a carboxylic acid pairing up in benzene.

Concept 1 of 3: Ranking solutions by particle concentration, i × m

Every colligative effect counts particles. So to rank solutions by boiling point, freezing point or osmotic pressure, work out

i \times

concentration for each and sort. A salt at a lower concentration can still beat a non-electrolyte at a higher one.

Definition

Effective concentration $= i \times m$ (or $i \times C$ ). The largest has the highest boiling point and the LOWEST freezing point.
Complete dissociation: glucose, urea 1; NaCl, KI 2; $\mathrm{CaCl_2}$ , $\mathrm{K_2SO_4}$ 3; $\mathrm{AlCl_3}$ 4; $\mathrm{Al_2(SO_4)_3}$ 5.
$\mathrm{KHSO_4}$ gives $\mathrm{K^+}$ and $\mathrm{HSO_4^-}$ , which ionises further, so its $i$ lies between 2 and 3.
A strong electrolyte's $i$ rises towards its full ion count on dilution, because the ions attract each other less when far apart.
When the solutions differ in both salt and concentration, compute each $i \times m$ ; do not rank by $i$ alone.

Colligative effects scale with particle concentration

\Delta T_b = i K_b m \qquad \Delta T_f = i K_f m \qquad \pi = iCRT

Worked example

Arrange in increasing order of freezing point, assuming complete dissociation: 0.1 m glucose, 0.05 m

\mathrm{K_2SO_4}

, 0.04 m

\mathrm{AlCl_3}

, 0.06 m NaCl.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q39Moderate

Example 1 · Solutions · Van't Hoff Factor and Abnormal Molar Mass

Consider the following aqueous solutions. (I) 2.2 g Glucose in 125 mL of solution. (II) 1.9 g Calcium chloride in 250 mL of solution. (III) 9.0 g Urea in 500 mL of solution. (IV) 20.5 g Aluminum sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in

gmol^{- 1}:H = 1,C = 12,\text{ }N = 14

O = 16,Cl = 35.5,Ca = 40,Al = 27

and

S = 32

]

Ten times the concentration beats twice the ions

0.01 M KCl has

iC = 0.02

; 0.001 M KCl has

0.002

. Two solutions of the same salt are not about equal when one is ten times as concentrated. Compute

i \times C

for each.

Dilution raises a strong electrolyte's i

For NaCl at 0.1, 0.01 and 0.001 M,

i

increases towards 2 as the solution gets more dilute. The reversed order is the planted option.

Drill 6 more on ranking solutions by particle concentration, i × m

Concept 2 of 3: Van 't Hoff factor for dissociation, i = 1 + (n − 1)α

Start with one mole. A fraction

\alpha

splits into

n

ions and the rest stays whole, so the particles are

1 - \alpha + n\alpha = 1 + (n - 1)\alpha

. A measured shift divided by the expected shift gives

i

, and

i

gives

\alpha

Definition

$i = \dfrac{\text{observed shift}}{\text{expected shift}} = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}$ .
Dissociation into $n$ ions: $i = 1 + (n - 1)\alpha$ , so $\alpha = \dfrac{i - 1}{n - 1}$ . For $\mathrm{MX_2}$ or $\mathrm{A_2B}$ , $n = 3$ ; for $\mathrm{MX_3}$ , $n = 4$ .
Weak monobasic acid HA: $i = 1 + \alpha$ , $K_a = \dfrac{C\alpha^2}{1 - \alpha}$ , and $\alpha \approx \sqrt{K_a/C}$ when $\alpha$ is small.
The observed shift is $(i - 1) \times 100\%$ above the expected one: $\alpha = 0.2$ for HA makes it 20% larger.
A precipitate removes its ions from solution, so recount the particles after it forms.

Degree of dissociation

i = 1 + (n - 1)\alpha \qquad \alpha = \frac{i - 1}{n - 1} \qquad K_a = \frac{C\alpha^2}{1 - \alpha}

Worked example

A 0.2 m aqueous solution of

\mathrm{AB_2}

(

\mathrm{AB_2 \rightarrow A^{2+} + 2B^-}

) freezes at

-0.93

°C. Find

i

and the degree of dissociation (

K_f = 1.86

K kg/mol).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q30Moderate

Example 2 · Solutions · Van't Hoff Factor and Abnormal Molar Mass

19.5 g of fluoro acetic acid (molar mass

= 78\text{ }g{\text{ }mol}^{- 1}

) is dissolved in 500 g of water at 298 K . The depression in the freezing point of water was

1^{\circ}C

. What is

K_{a}

of fluoro acetic acid ? (For water,

K_{f}= 1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1}

). Assume molarity and molality to have same values.

Divide by n − 1, not by n

\alpha = (i - 1)/(n - 1)

. For

\mathrm{MX_3}

with

i = 1.9

\alpha = 0.9/3 = 0.3

, not

0.9/4

. For a two-ion salt it is simply

i - 1

Count the ions from the formula

\mathrm{MX_2}

and

\mathrm{A_2B}

give three ions each;

\mathrm{MX_3}

gives four. Using

n = 2

for every salt gives the wrong degree of dissociation.

Observed molar mass is LOWER for dissociation

More particles mean a larger shift and so a smaller apparent molar mass:

M_{\text{obs}} = M_{\text{normal}}/i

. If your observed mass came out larger than the formula mass for a salt, the ratio is upside down.

Drill 15 more on van 't hoff factor for dissociation, i = 1 + (n − 1)α

Concept 3 of 3: Van 't Hoff factor for association, i = 1 − (1 − 1/n)α

When molecules pair up, two become one particle, so the count falls. Carboxylic acids do this in benzene: two molecules hold each other by two hydrogen bonds. The shift is smaller than expected and the apparent molar mass is larger.

Definition

Association into $n$ -mers: $i = 1 - \alpha + \dfrac{\alpha}{n} = 1 - \left(1 - \dfrac{1}{n}\right)\alpha$ .
Dimers ( $n = 2$ ): $i = 1 - \dfrac{\alpha}{2}$ , so $\alpha = 2(1 - i)$ .
Complete association: $i = 1/n$ . Benzoic acid and acetic acid in benzene give $i \approx 0.5$ : dimers.
Observed molar mass $= M/i$ , larger than the formula mass.
Mixed fates add: if 40% of HA dimerises and the other 60% splits into two ions, $i = 0.2 + 1.2 = 1.4$ .

Degree of association

i = 1 - \left(1 - \frac{1}{n}\right)\alpha \qquad \text{dimer: } i = 1 - \frac{\alpha}{2}

Worked example

1.2 g of acetic acid (M = 60) in 50 g of benzene lowers its freezing point by 1.28 K (

K_f = 5.12

K kg/mol). Find the percent association into dimers and the observed molar mass.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q123Moderate

Example 3 · Solutions · Van't Hoff Factor and Abnormal Molar Mass

Solute

A

associates in water. When

0.7\text{ }g

of solute

A

is dissolved in

42.0\text{ }g

of water, it depresses the freezing point by

0.2^{\circ}C

. The percentage association of solute

A

in water, is [Given: Molar mass of

A = 93\text{ }g{\text{ }mol}^{- 1}

. Molal depression constant of water is

1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1}

]

A dimer halves, it does not vanish

For dimerisation

i = 1 - \alpha/2

, not

1 - \alpha

. With

i = 0.7

\alpha = 0.6

; using

1 - \alpha

gives 0.3 and a wrong percentage.

Association raises the apparent molar mass

Fewer particles give a smaller shift, so the molar mass worked out from it is too LARGE: about twice the formula mass for a fully dimerised acid.

Drill 4 more on van 't hoff factor for association, i = 1 − (1 − 1/n)α

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Ranking solutions by particle concentration, i × m
Colligative effects scale with particle concentration
$\Delta T_b = i K_b m \qquad \Delta T_f = i K_f m \qquad \pi = iCRT$
Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
Degree of dissociation
$i = 1 + (n - 1)\alpha \qquad \alpha = \frac{i - 1}{n - 1} \qquad K_a = \frac{C\alpha^2}{1 - \alpha}$
Van 't Hoff factor for association, i = 1 − (1 − 1/n)α
Degree of association
$i = 1 - \left(1 - \frac{1}{n}\right)\alpha \qquad \text{dimer: } i = 1 - \frac{\alpha}{2}$

Watch out for (7)

Ten times the concentration beats twice the ions
→ Ranking solutions by particle concentration, i × m
Dilution raises a strong electrolyte's i
→ Ranking solutions by particle concentration, i × m
Divide by n − 1, not by n
→ Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
Count the ions from the formula
→ Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
Observed molar mass is LOWER for dissociation
→ Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
A dimer halves, it does not vanish
→ Van 't Hoff factor for association, i = 1 − (1 − 1/n)α
Association raises the apparent molar mass
→ Van 't Hoff factor for association, i = 1 − (1 − 1/n)α

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