MHT-CET Chemistry · Solutions and Colligative Properties

Van't Hoff Factor and Abnormal Molar Mass

i = (observed colligative property)/(calculated for no dissociation) = ΔTf/(Kf·m); i > 1 for dissociation, i < 1 for association, and the degree of dissociation is α = (i − 1)/(n − 1).

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Why this matters

25 PYQs, three HARD — the page that makes the electrolyte stems on every other page honest. Two-thirds are i from a measured freezing point of a 0.15–0.2 m KCl-type solution (i comes out near 1.8), the rest are the 'x K for urea, so how much for CaCl₂' ratio (3x) or AlCl₃ (4x), one percent-dissociation calculation, and the recall of which properties are colligative. Learn the ion counts and the two rearrangements; the arithmetic is one division.

Concept 1 of 3: The Van't Hoff Factor: i = ΔTf(observed)/(Kf · m), and the Ion Count for Complete Dissociation

An electrolyte gives more particles than its formula count, so its colligative effect is larger than the molality predicts. The ratio of observed to expected is

i

; a fully dissociated salt has

i

equal to its number of ions, a real solution a little less.

Definition

$i = \dfrac{\Delta T_f^{\text{obs}}}{K_f\,m}$ : $0.2$ m freezing at $-0.68^\circ$ C gives $\dfrac{0.68}{1.86 \times 0.2} = 1.83$ ; $0.15$ m at $-0.51^\circ$ C gives $1.82$ ; $0.2$ m at $-0.7$ K gives $1.88$ ; $0.18$ m at $-0.54^\circ$ C gives $1.61$ ; $0.01$ m at $-0.056$ K gives $3.00$ .
Complete dissociation: KCl, NaCl $\to 2$ ; CaCl $_2$ , BaCl $_2$ , Na $_2$ SO $_4$ $\to 3$ ; AlCl $_3$ $\to 4$ ; Al $_2$ (SO $_4$ ) $_3$ $\to 5$ . Non-electrolytes (urea, glucose, sucrose) $i = 1$ .
Ratio stems: if $1$ m urea gives $\Delta T_f = x$ , $1$ m CaCl $_2$ gives $3x$ ; if a non-electrolyte gives $\Delta T_b = x$ , AlCl $_3$ at the same molality gives $4x$ .
$\Delta T_f = i K_f m$ forwards: $0.01$ m formic acid with $i = 1.1$ gives $1.1 \times 0.01 \times 1.86 = 0.020$ K.
Association (acetic acid dimerising in benzene) gives $i < 1$ and an abnormally HIGH observed molar mass; dissociation gives an abnormally LOW one.

Van't Hoff factor

i = \frac{\text{observed}}{\text{calculated}} = \frac{\Delta T_f^{\text{obs}}}{K_f\,m} = \frac{M_{\text{calc}}}{M_{\text{obs}}}

Worked example

0.1

m aqueous solution of an electrolyte freezes at

-0.465^\circ

C. Find

i

and suggest the number of ions it gives.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q86Moderate

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

0.2 molal aqueous solution of KCl freezes at

-0.680^\circ\text{C}

. Calculate van't Hoff factor for this solution.

K_f=1.86\,\text{K kg mol}^{-1}

Using the calculated ΔTf as the observed one

K_f m = 0.372

is the EXPECTED depression for

0.2

m; the observed

0.68

goes on top. Inverting gives

0.55

, option (D) on the

-0.660

K stem.

Drill 10 more on the van't hoff factor: i = δtf(observed)/(kf · m), and the ion count for complete dissociation

Concept 2 of 3: Degree of Dissociation From i: α = (i − 1)/(n − 1)

If a fraction

\alpha

of the formula units split into

n

ions, one mole becomes

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

moles of particles. That is

i

, so

\alpha = \dfrac{i - 1}{n - 1}

. For association into

n

-mers,

i = 1 - \alpha + \dfrac{\alpha}{n}

Definition

$0.02$ m, $\Delta T_f = 0.046$ K, $n = 2$ : $i = \dfrac{0.046}{0.0372} = 1.236$ , $\alpha = 0.236$ , i.e. $23.6\%$ dissociation.
KCl with $i = 1.83$ : $\alpha = 0.83$ . A salt of three ions with $i = 2.5$ : $\alpha = 0.75$ .
Observed molar mass $= \dfrac{M_{\text{calc}}}{i}$ : smaller than the formula mass for dissociation, larger for association.

Degree of dissociation

i = 1 + (n - 1)\alpha \ \Rightarrow\ \alpha = \frac{i - 1}{n - 1} \qquad \text{association: } i = 1 - \alpha + \frac{\alpha}{n}

Worked example

0.1

m solution of a salt MX

_2

freezes at

-0.4092^\circ

C. Find its percent dissociation.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q54Moderate

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

Calculate the percent dissociation of 0.02 m solution if its freezing point depression is 0.046 K .

\left\lbrack K_{f} \right.

for water

\left. \ = 1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1};n = 2 \right\rbrack

Dividing by n instead of n − 1

\alpha = \dfrac{i - 1}{n - 1}

; for

n = 2

that is simply

i - 1

. Dividing

0.236

2

gives

11.8\%

, close to option (A).

Concept 3 of 3: Which Properties Are Colligative, and the Statements the Exam Tests

A colligative property depends only on the NUMBER of solute particles, not their identity. There are exactly four; a boiling point on its own is a property of the pure liquid and is not one of them.

Definition

The four: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure.
NOT colligative: boiling point, freezing point, vapour pressure themselves (they are the solvent's properties), osmosis (a process, not a property), density, viscosity.
True statements: a non-volatile solute LOWERS the vapour pressure and RAISES the boiling point; $0.1$ M NaCl has a HIGHER osmotic pressure than $0.1$ M sucrose ( $i \approx 2$ ); $\Delta T_f = K_f$ numerically for a $1$ molal solution.

Property	Colligative?	Formula
Relative lowering of vapour pressure	Yes	$\dfrac{P^\circ - P}{P^\circ} = x_2$
Elevation of boiling point	Yes	$\Delta T_b = i K_b m$
Depression of freezing point	Yes	$\Delta T_f = i K_f m$
Osmotic pressure	Yes	$\pi = i C R T$
Boiling point (of the solution)	No	An intensive property of the liquid; its CHANGE is colligative 'Boiling point' alone is the standard wrong answer to 'which is not colligative'.
Osmosis	No	A process; osmotic PRESSURE is the property 'Osmosis is a colligative property' is a planted false statement.

Four properties, all proportional to particle count; the quantities they change are not themselves colligative.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q70Moderate

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

Which among the following is not colligative property?

Marking 'boiling point elevation' as the non-colligative one

The option list mixes 'boiling point' with 'freezing point depression'. The bare property is the odd one out; anything with 'elevation', 'depression' or 'lowering' in its name is colligative.

Drill 2 more on which properties are colligative, and the statements the exam tests

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 (2)

The Van't Hoff Factor: i = ΔTf(observed)/(Kf · m), and the Ion Count for Complete Dissociation
Van't Hoff factor
$i = \frac{\text{observed}}{\text{calculated}} = \frac{\Delta T_f^{\text{obs}}}{K_f\,m} = \frac{M_{\text{calc}}}{M_{\text{obs}}}$
Degree of Dissociation From i: α = (i − 1)/(n − 1)
Degree of dissociation
$i = 1 + (n - 1)\alpha \ \Rightarrow\ \alpha = \frac{i - 1}{n - 1} \qquad \text{association: } i = 1 - \alpha + \frac{\alpha}{n}$

Reference tables (1)

Which Properties Are Colligative, and the Statements the Exam Tests6 rows

Property	Colligative?	Formula
Relative lowering of vapour pressure	Yes	$\dfrac{P^\circ - P}{P^\circ} = x_2$
Elevation of boiling point	Yes	$\Delta T_b = i K_b m$
Depression of freezing point	Yes	$\Delta T_f = i K_f m$
Osmotic pressure	Yes	$\pi = i C R T$
Boiling point (of the solution)	No	An intensive property of the liquid; its CHANGE is colligative 'Boiling point' alone is the standard wrong answer to 'which is not colligative'.
Osmosis	No	A process; osmotic PRESSURE is the property 'Osmosis is a colligative property' is a planted false statement.

Four properties, all proportional to particle count; the quantities they change are not themselves colligative.

Watch out for (3)

Using the calculated ΔTf as the observed one
→ The Van't Hoff Factor: i = ΔTf(observed)/(Kf · m), and the Ion Count for Complete Dissociation
Dividing by n instead of n − 1
→ Degree of Dissociation From i: α = (i − 1)/(n − 1)
Marking 'boiling point elevation' as the non-colligative one
→ Which Properties Are Colligative, and the Statements the Exam Tests

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Related notes

Elevation of Boiling Point — the i × m rankings →