We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001

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Q. We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

(i)

i_A < i_B < i_C

(ii)

i_A > i_B > i_C

(iii)

i_A = i_B = i_C

(iv)

i_A < i_B > i_C

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Concept understanding — VanT Hoff Factor Association

The Intuition: What Happens When Particles Stick Together?

Imagine you're counting people in a room. You see 100 chairs, each with one person. That's 100 individuals. Now imagine those same 100 people decide to pair up — every two people hold hands and become a "couple." Suddenly, the number of independent moving units in the room drops from 100 to 50.

That's exactly what association does in a solution. When solute particles (molecules or ions) associate, they clump together into larger clusters. The number of independent particles floating around decreases. And since colligative properties (freezing point depression, boiling point elevation, osmotic pressure) depend only on the number of particles — not their identity — the observed effect becomes smaller than expected.

Note

Association is the opposite of dissociation. In dissociation, one particle breaks into many (e.g., NaCl → Na⁺ + Cl⁻). In association, many particles combine into one (e.g., two acetic acid molecules dimerise).

The Van't Hoff Factor: The Correction Number

The Van't Hoff factor, denoted by $i$ , is defined as:

i = \frac{\text{Actual number of particles in solution after association}}{\text{Number of particles if no association occurred}}

For a non-electrolyte that does not associate or dissociate, $i = 1$ .

For association, $i < 1$ — because the actual particle count is less than what you started with.

A Concrete Example: Acetic Acid in Benzene

Acetic acid ( $\text{CH}_3\text{COOH}$ ) in benzene forms dimers — two molecules stick together via hydrogen bonding:

2\,\text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2

Suppose you dissolve 100 molecules of acetic acid. If no association occurred, you'd have 100 particles. But if all of them dimerise, you get only 50 dimers. So:

i = \frac{50}{100} = 0.5

In reality, association is never 100% complete — it's an equilibrium. So $i$ lies between 0.5 and 1.

Watch out

A common mistake: thinking $i$ can be negative. It cannot. For association, $0 < i < 1$ . For dissociation, $i > 1$ . For no change, $i = 1$ .

The General Formula for Association

Let’s say $n$ molecules of a solute associate to form one associated particle:

nA \rightleftharpoons A_n

Let $\alpha$ be the degree of association — the fraction of original molecules that have associated.

Initially: 1 mole of A (i.e., $N$ molecules)
Moles that associate: $\alpha$
Moles that remain as single A: $1 - \alpha$
Moles of associated particles formed: $\frac{\alpha}{n}$ (because $n$ molecules make 1 associated unit)

Total moles after association:

(1 - \alpha) + \frac{\alpha}{n}

The Van't Hoff factor is:

i = \frac{\text{Total moles after association}}{\text{Initial moles}} = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1} = 1 - \alpha + \frac{\alpha}{n}

Simplify:

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

For the common case of dimerisation ( $n = 2$ ):

i = 1 - \alpha\left(1 - \frac{1}{2}\right) = 1 - \frac{\alpha}{2}

So if $\alpha = 0.6$ (60% association), then $i = 1 - 0.3 = 0.7$ .

How Association Affects Colligative Properties

All colligative properties are multiplied by $i$ :

Property	Formula without association	Formula with association
Relative lowering of vapour pressure	$\frac{p^\circ - p}{p^\circ} = x_B$	$\frac{p^\circ - p}{p^\circ} = i \cdot x_B$
Elevation in boiling point	$\Delta T_b = K_b \cdot m$	$\Delta T_b = i \cdot K_b \cdot m$

Why this formula?

Van't Hoff Factor for Association: Why the Formula Holds

The Van't Hoff factor ( $i$ ) for association describes how solute particles combine in solution, reducing the effective number of particles. Let's build the reasoning step-by-step.

1. The Core Idea: What Changes?

When a solute associates (e.g., two acetic acid molecules dimerize in benzene), the number of particles in solution decreases. The Van't Hoff factor is defined as:

i = \frac{\text{Actual number of particles in solution}}{\text{Number of particles if no association}}

For association, $i < 1$ .

2. Setting Up the Association Process

Consider a solute that associates to form $n$ molecules per aggregate (e.g., $n=2$ for dimerization). Let:

Initial moles of solute = $1$ mole (for simplicity)
Degree of association = $\alpha$ (fraction of solute that associates)

What happens to the particles?

Moles that associate = $\alpha$ (these combine into aggregates)
Moles that remain free = $1 - \alpha$

Each associated group of $n$ molecules becomes 1 aggregate particle. So:

Number of aggregates formed = $\frac{\alpha}{n}$
Number of free molecules = $1 - \alpha$

3. Total Particles After Association

Total moles of particles in solution:

\text{Total} = \underbrace{(1 - \alpha)}_{\text{free}} + \underbrace{\frac{\alpha}{n}}_{\text{aggregates}}

If no association ( $\alpha = 0$ ), total = $1$ mole of particles.

4. The Van't Hoff Factor Formula

By definition:

i = \frac{\text{Total particles after association}}{\text{Total particles if no association}} = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1}

Thus:

\boxed{i = 1 - \alpha + \frac{\alpha}{n}}

5. Why This Makes Physical Sense

If $\alpha = 0$ (no association): $i = 1$ — particles behave independently.
If $\alpha = 1$ (complete association): $i = \frac{1}{n}$ — all molecules form $n$ -mers, so particle count drops by factor $n$ . …

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The van’t Hoff factor $i$ for NaCl increases as the solution becomes more dilute because fewer ion pairs form. So the order is $i_A < i_B < i_C$ , which corresponds to option (i).

Why the van’t Hoff factor depends on concentration

NaCl is a strong electrolyte — in theory it dissociates completely into Na⁺ and Cl⁻, giving $i = 2$ . But in reality, at higher concentrations, some ions come close enough to feel electrostatic attraction and temporarily pair up (ion pairing). This reduces the effective number of particles. The effect is strongest when the solution is concentrated, and it weakens as we dilute.

So the van’t Hoff factor is not a constant for a given solute — it approaches the theoretical value only in the limit of infinite dilution. For NaCl, $i$ is always slightly less than 2, and it gets closer to 2 as the concentration drops.

Watch out

A common mistake is to assume $i = 2$ for all concentrations of a strong electrolyte. That’s only true at infinite dilution. In real solutions, ion pairing lowers $i$ , especially at higher molarities.

Step-by-step reasoning

Recall the definition The van’t Hoff factor $i$ is the ratio of the actual number of particles in solution to the number of formula units dissolved. For NaCl, if dissociation were complete:

i = \frac{\text{moles of particles}}{\text{moles of NaCl}} = 2

Identify the real behaviour

At finite concentrations, some Na⁺ and Cl⁻ ions associate transiently into ion pairs (Na⁺Cl⁻). These pairs count as one particle, not two. So the actual particle count is less than $2n$ , meaning $i < 2$ .
Connect concentration to ion pairing …

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Common Mistakes Students Make on van't Hoff Factor & Concentration

Mistake 1: Assuming $i$ is constant for all concentrations of the same solute

Why it happens: Students memorize that for NaCl, $i = 2$ (complete dissociation into Na⁺ and Cl⁻) and apply it blindly.

The truth: The van't Hoff factor $i$ depends on degree of dissociation, which changes with concentration. At infinite dilution, dissociation is complete ( $i \to 2$ ). At higher concentrations, ion pairing reduces $i$ .

How to avoid: Always ask: "Is this at infinite dilution or a real concentration?" For real solutions, $i$ decreases as concentration increases.

Mistake 2: Thinking higher concentration means higher $i$

Why it happens: Students confuse "more ions present" with "higher $i$ ". More concentrated solutions do have more ions per litre, but $i$ is a ratio:

i = \frac{\text{actual number of particles}}{\text{number of formula units dissolved}}

The truth: At higher concentration, ions are closer together and more likely to recombine (ion pairing), so the degree of dissociation decreases, making $i$ smaller.

How to avoid: Remember: Dilution favours dissociation. As concentration decreases, $i$ approaches the theoretical maximum.

Mistake 3: Picking option (iii) — $i_A = i_B = i_C$

Why it happens: Students think "same solute, same $i$ ".

The truth: $i$ is concentration-dependent. For NaCl:

At 0.1 M: significant ion pairing → $i \approx 1.87$
At 0.01 M: less pairing → $i \approx 1.94$
At 0.001 M: nearly complete dissociation → $i \approx 1.99$

How to avoid: Visualise the trend: more dilute → more dissociation → higher $i$ .

Mistake 4: Picking option (ii) — $i_A > i_B > i_C$

Why it happens: Students think "more concentrated means more ions, so higher $i$ ".

The truth: This reverses the actual trend. The correct order is:

\boxed{i_A < i_B < i_C}

which corresponds to option (i). …

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Q.We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

Concept understanding — VanT Hoff Factor Association

The Intuition: What Happens When Particles Stick Together?

The Van't Hoff Factor: The Correction Number

A Concrete Example: Acetic Acid in Benzene

The General Formula for Association

How Association Affects Colligative Properties

Why this formula?

Van't Hoff Factor for Association: Why the Formula Holds

1. The Core Idea: What Changes?

2. Setting Up the Association Process

What happens to the particles?

3. Total Particles After Association

4. The Van't Hoff Factor Formula

5. Why This Makes Physical Sense

Why the van’t Hoff factor depends on concentration

Step-by-step reasoning

Method: Effect of Dilution on van’t Hoff Factor for Strong Electrolytes

Steps to determine the order:

Common Mistakes Students Make on van't Hoff Factor & Concentration

Mistake 1: Assuming $i$ is constant for all concentrations of the same solute

Mistake 2: Thinking higher concentration means higher $i$

Mistake 3: Picking option (iii) — $i_A = i_B = i_C$

Mistake 4: Picking option (ii) — $i_A > i_B > i_C$

Q.We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

Why the van’t Hoff factor depends on concentration

Step-by-step reasoning

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