0.6mol of NaCl is dissolved in 1kg of water. Assuming NaCl dissociates completely into its ions, calculate the

Example · Example 18

0.6\ \text{mol}

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Concept understanding — Van't Hoff Factor

The Intuition: Why a Single Solute Can Act Like Many

Imagine you drop a sugar cube into a cup of tea. The sugar molecules stay whole — each molecule remains one particle. Now imagine you drop a pinch of table salt (NaCl) into water. The salt doesn't stay as NaCl molecules; it splits apart into Na⁺ and Cl⁻ ions. One formula unit of NaCl becomes two separate particles in solution.

If you measure the boiling point elevation or freezing point depression of these solutions, the salt solution behaves as if it has twice as many solute particles as the sugar solution — even though you dissolved the same number of formula units. That's the core idea: colligative properties depend on the number of particles, not the identity of the particles.

But what about substances that do the opposite? Acetic acid in benzene, for example, can pair up into dimers — two molecules associating to form one effective particle. That reduces the particle count.

The Van't Hoff factor $i$ is simply the number that tells you: for every one formula unit you dissolve, how many particles actually end up floating around in solution?

The Precise Definition

i = \frac{\text{observed value of a colligative property}}{\text{expected value if no dissociation or association occurs}}

For an ideal, non-dissociating, non-associating solute, $i = 1$ . For a solute that dissociates completely, $i$ equals the number of ions produced per formula unit. For a solute that associates, $i$ is less than 1.

How It Connects to Colligative Properties

Every colligative property formula gets multiplied by $i$ :

Relative lowering of vapour pressure: $\frac{p^\circ - p}{p^\circ} = i \cdot \frac{n_2}{n_1}$
Elevation of boiling point: $\Delta T_b = i \cdot K_b \cdot m$
Depression of freezing point: $\Delta T_f = i \cdot K_f \cdot m$
Osmotic pressure: $\Pi = i \cdot CRT$

Where $m$ is molality, $C$ is molar concentration, and $n_1$ , $n_2$ are moles of solvent and solute respectively.

The Two Cases in Detail

Dissociation (i > 1)

When a solute breaks into ions, the number of particles increases. For a compound $A_xB_y$ that dissociates completely:

A_xB_y \rightarrow xA^{y+} + yB^{x-}

The maximum possible $i$ is $x + y$ . But if dissociation is partial, $i$ lies between 1 and $x+y$ .

The degree of dissociation $\alpha$ relates to $i$ by:

i = 1 + (n-1)\alpha

where $n$ is the number of ions produced from one formula unit. For example, for NaCl ( $n=2$ ), if $\alpha = 0.8$ , then $i = 1 + (2-1)(0.8) = 1.8$ .

Tip

For complete dissociation of NaCl, $i = 2$ . For CaCl₂, $i = 3$ . For Al₂(SO₄)₃, $i = 5$ . Memorise these common ones — they appear frequently in numerical problems.

Association (i < 1)

When solute molecules combine to form larger aggregates, the particle count drops. For a dimerisation:

2A \rightleftharpoons A_2

If $n$ molecules associate to form one aggregate, and the degree of association is $\alpha$ :

i = 1 - \left(1 - \frac{1}{n}\right)\alpha

For dimerisation ( $n=2$ ), this becomes $i = 1 - \frac{\alpha}{2}$ .

Watch out

A common mistake: assuming $i$ is always greater than 1. It is not. Association in non-polar solvents (like carboxylic acids in benzene) gives $i < 1$ . Always check whether the solute is likely to dissociate or associate in the given solvent.

A Worked Example …

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Since $\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$ dissociates completely into 2 ions per formula unit, the van't Hoff factor is $i=2$ . With molality $m = 0.6\ \text{mol kg}^{-1}$ , the corrected freezing-point depression is $\Delta T_f = i K_f m = 2 \times 1.86 \times 0.6 = 2.232\ \text{K}$ . [!ANSWER] The observed fr …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set V11 mark
$Q. Van't Hoff factor$
(i) for a non-electrolyte in a solution is ________.
›Reveal solutionSolution

For a non-electrolyte solute the van't Hoff factor $i = 1$ , since the number of particles in solution is unchanged.

The van't Hoff factor is

$i = \dfrac{\text{actual number of particles in solution after dissociation/association}}{\text{number of formula units dissolved}}$ …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The strong electrolyte having maximum value of Van't Hoff factor$
(i) among the following is
(a) NaCl
(b) KCl
(c) MgSO4
(d) K2SO4
›Reveal solutionSolution
The van't Hoff factor i equals the number of ions a formula unit produces on complete dissociation, so the electrolyte giving the most ions has the largest i.

NaCl -> Na+ + Cl- : i = 2

KCl -> K+ + Cl- : i = 2

MgSO4 -> Mg2+ + SO4^2- : i = 2

K2SO4 -> 2K+ + SO4^2- : i = 3 …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the value of Vant Hoff factor for one molal aqueous solution of Urea?$
(a) Two
(b) Three
(c) One
(d) Four
›Reveal solutionSolution

Van't Hoff factor $i$ measures how many particles a solute actually produces in solution. Urea is a non-electrolyte that neither dissociates nor associates, so $i = 1$ .

The Van't Hoff factor is defined as

$i = \dfrac{\text{observed colligative property}}{\text{calculated (normal) colligative property}} = \dfrac{\text{actual number of particles after dissolution}}{\text{number of formula units dissolved}}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. When mercuric iodide is added to the aqueous solution of Potassium iodide:$
(a) Freezing point is raised
(b) Freezing point does not change
(c) Freezing point is lowered
(d) Boiling point does not change
›Reveal solutionSolution

$HgI_2$ combines with excess $I^-$ to form the complex ion $[HgI_4]^{2-}$ , which decreases the number of particles in solution, so the depression in freezing point is less than expected and the freezing point is raised.

When mercuric iodide is added to an aqueous solution of potassium iodide, it reacts with the iodide ions already present:

$HgI_2 + 2I^- \rightarrow [HgI_4]^{2-}$ …

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CBSE 2026Set ANNUAL1 mark
$Q. Write answer in one word/sentence: Write the numerical value of 'i' for complete dissociation of electrolyte K2SO4.$
›Reveal solutionSolution

Complete dissociation of K2SO4 gives 3 ions, so i = 3.

The van't Hoff factor i is the number of particles actually present in solution per formula unit of the dissolved solute. On complete dissociation:

K2SO4 -> 2K+ + SO4^2-

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one of the following has maximum value of Van't Hoff Factor -$
(a) K2SO4
(b) NaCl
(c) MgSO4
(d) KCl
›Reveal solutionSolution
Van't Hoff factor increases with the number of ions produced on complete dissociation.

For a strong electrolyte dissociating completely into $n$ ions, the Van't Hoff factor $i \approx n$ .

$K_2SO_4 \to 2K^+ + SO_4^{2-}$ : $n = 3$

$NaCl \to Na^+ + Cl^-$ : $n = 2$

$MgSO_4 \to Mg^{2+} + SO_4^{2-}$ : $n = 2$ …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of Van't Hoff factor$
(i) for complete dissociation of MgSO4 is -
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

MgSO4 is a strong electrolyte that on complete dissociation gives 2 ions per formula unit, so i = 2.

The Van't Hoff factor is defined as:

i = (observed colligative property)/(calculated colligative property assuming no dissociation) = (actual number of particles in solution)/(number of formula units dissolved)

MgSO4 dissociates in water as:

MgSO4 -> Mg2+ + SO4^2-

One formula unit gives 2 ions (1 Mg2+ + 1 SO4^2-) on complete dissociation. So for complete dissociation: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For solutes which do not undergo any association or dissociation in solution, the Van't Hoff factor$
(i) will be –
(i) less than 1
(ii) more than 1
(iii) equal to 1
(iv) zero
›Reveal solutionSolution
The van't Hoff factor $i$ measures the ratio of actual to normal number of particles; if a solute neither associates nor dissociates, that ratio is exactly 1.

The van't Hoff factor is defined as:

$i = \dfrac{\text{normal molar mass}}{\text{observed molar mass}} = \dfrac{\text{observed number of particles after dissolution}}{\text{number of particles as per formula}}$

If the solute dissociates (e.g. NaCl → Na⁺ + Cl⁻), the number of particles increases, so $i > 1$ . …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. In comparison to a 0.01 M solution of glucose, the depression in the freezing point of 0.01 M MgCl2 solution is :$
(a) remains same
(b) about twice
(c) about three times
(d) about four times
›Reveal solutionSolution

$MgCl_2$ dissociates into 3 ions (van't Hoff factor $i \approx 3$ ), so its colligative effect — including freezing-point depression — is about 3 times that of a non-electrolyte like glucose at the same molar concentration.

Depression in freezing point is a colligative property proportional to the number of solute particles in solution:

$\Delta T_f = i \cdot K_f \cdot m$

Glucose is a non-electrolyte, so $i = 1$ (it does not dissociate). …

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CBSE 2024Set ANNUAL1 mark
$Q. What is Van't Hoff factor? OR Define Activation Energy.$
›Reveal solutionSolution
Van't Hoff factor i measures how many particles a solute actually produces in solution compared to the 'ideal' assumption of one particle per formula unit.

Raoult's law and the colligative-property formulas (boiling-point elevation, freezing-point depression, osmotic pressure, etc.) assume the solute stays as single, non-interacting particles. But some solutes DISSOCIATE into more particles (e.g. NaCl → Na⁺ + Cl⁻, doubling the particle count) while others ASSOCIATE into fewer, larger particles (e.g. some carboxylic acids dimerise in non-polar solvents). This causes the OBSERVED colligative property to differ from the value calculated for a simple, non-dissociating solute.

The Van't Hoff factor is defined as:

i = (observed colligative property) / (calculated colligative property for no dissociation)

Equivalently, i = actual number of particles in solution after dissociation/association ÷ number of formula units initially dissolved.

i > 1 → the solute dissociates (more particles than expected) …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. When benzoic acid is dissolved in benzene, the van't Hoff factor will be$
(a) 1
(b) 0.5
(c) 2
(d) 1.5
›Reveal solutionSolution

Benzoic acid dimerises in benzene through intermolecular H-bonding, so the number of effective particles halves and $i = 1/n = 0.5$ .

The van't Hoff factor $i$ measures how the effective number of particles in solution deviates from the number of formula units dissolved:

$i = \dfrac{\text{normal molar mass}}{\text{observed (abnormal) molar mass}} = \dfrac{\text{observed colligative property}}{\text{calculated colligative property (no association/dissociation)}}$

In a polar, protic solvent (e.g. water), benzoic acid partially ionises, $i>1$ . But benzene is non-polar, so ionisation cannot occur. Instead, two molecules of benzoic acid associate through two intermolecular hydrogen bonds between the $-COOH$ groups of each molecule, forming a stable cyclic dimer:

$2\,C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$

If association is complete, every $n=2$ monomer units behave as a single kinetic particle, so

$i = \dfrac{1}{n} = \dfrac{1}{2} = 0.5$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The values of Van't Hoff factors for KCl, NaCl, K2SO4 respectively are$
(a) 2, 2 and 2
(b) 2, 2 and 3
(c) 1, 1 and 2
(d) 1, 1 and 1
›Reveal solutionSolution

Van't Hoff factor for a fully-dissociating strong electrolyte equals the number of ions it produces per formula unit.

KCl → K⁺ + Cl⁻ gives 2 ions ⟹ i = 2. NaCl → Na⁺ + Cl⁻ gives 2 ions ⟹ i = 2. K₂SO₄ → 2K⁺ + SO₄²⁻ gives 3 ions ⟹ i = 3. (These are the theoretical/limiting values assuming complete dissociation; measured i i …

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