The values of Van't Hoff factors for KCl, NaCl and $K_2SO_4$, respectively, are _____________. (i) 2, 2 and 2 (ii) 2, 2 and 3 (iii) 1, 1 and 2 (iv) 1, 1 and 1

> [!TLDR] > The Van't Hoff factor counts the number of particles each formula unit produces in solution. KCl → 2 ions, NaCl → 2 ions, K₂SO₄ → 3 ions, giving **i = 2, 2, and 3**. The Van't Hoff factor $i$ measures how many particles a single formula unit of solute produces when it dissolves. For colligative properties—those that depend only on particle count, not identity—this factor scales the effect. A non-electrolyte like glucose stays intact ($i = 1$), but electrolytes dissociate into ions, so $i$ equals the total number of ions per formula unit. The key is to write the dissociation equation for each salt and count the ions. ## Step-by-step dissociation 1. **Potassium chloride, KCl** When KCl dissolves in water, it separates completely into its constituent ions: $$\text{KCl} \longrightarrow \text{K}^+ + \text{Cl}^-$$ One formula unit yields **2 ions** (one potassium cation, one chloride anion). Therefore, $i_{\text{KCl}} = 2$. 2. **Sodium chloride, NaCl** Similarly, NaCl is a strong electrolyte that dissociates fully: $$\text{NaCl} \longrightarrow \text{Na}^+ + \text{Cl}^-$$ Again, **2 ions** per formula unit, so $i_{\text{NaCl}} = 2$. 3. **Potassium sulfate, K₂SO₄** This salt contains two potassium ions and one sulfate ion per formula unit. In solution: $$\text{K}_2\text{SO}_4 \longrightarrow 2\,\text{K}^+ + \text{SO}_4^{2-}$$ …

The values of Van't Hoff factors for KCl, NaCl and K₂SO₄, respectively, are _______. (i) 2, 2 and 2 (ii) 2, 2

NCERT Exemplar · Q15

K_2SO_4

(i) 2, 2 and 2

(ii) 2, 2 and 3

(iii) 1, 1 and 2

(iv) 1, 1 and 1

Puducherry CbseMCQ· 1mImportance★★★★★

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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 counts the number of particles each formula unit produces in solution. KCl → 2 ions, NaCl → 2 ions, K₂SO₄ → 3 ions, giving i = 2, 2, and 3.

The Van't Hoff factor $i$ measures how many particles a single formula unit of solute produces when it dissolves. For colligative properties—those that depend only on particle count, not identity—this factor scales the effect. A non-electrolyte like glucose stays intact ( $i = 1$ ), but electrolytes dissociate into ions, so $i$ equals the total number of ions per formula unit.

The key is to write the dissociation equation for each salt and count the ions.

Step-by-step dissociation

Potassium chloride, KCl When KCl dissolves in water, it separates completely into its constituent ions:

\text{KCl} \longrightarrow \text{K}^+ + \text{Cl}^-

One formula unit yields 2 ions (one potassium cation, one chloride anion). Therefore, $i_{\text{KCl}} = 2$ .

Sodium chloride, NaCl Similarly, NaCl is a strong electrolyte that dissociates fully:

\text{NaCl} \longrightarrow \text{Na}^+ + \text{Cl}^-

Again, 2 ions per formula unit, so $i_{\text{NaCl}} = 2$ .

Potassium sulfate, K₂SO₄ This salt contains two potassium ions and one sulfate ion per formula unit. In solution: $\text{K}_2\text{SO}_4 \longrightarrow 2\,\text{K}^+ + \text{SO}_4^{2-}$ …

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Method: Complete Dissociation Model for Strong Electrolytes

This method uses the van’t Hoff factor ( $i$ ) definition for strong electrolytes that dissociate completely in solution:

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

For a salt that dissociates into $n$ ions, $i = n$ (assuming 100% dissociation).

Steps

Write the dissociation equation for each salt.
- KCl:

\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-

 → 2 ions → $i = 2$

NaCl:

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

 → 2 ions → $i = 2$ …

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Q.The values of Van't Hoff factors for KCl, NaCl and $K_2SO_4$ , respectively, are _____________.

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

Step-by-step dissociation

Method: Complete Dissociation Model for Strong Electrolytes

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that KCl and NaCl are 1:1 electrolytes

Mistake 2: Assuming all salts have the same Van't Hoff factor

Mistake 3: Confusing association with dissociation

Mistake 4: Ignoring the charge on ions …

Q.The values of Van't Hoff factors for KCl, NaCl and K2SO4K_2SO_4K2​SO4​, respectively, are _____________.

Step-by-step dissociation

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Q.The values of Van't Hoff factors for KCl, NaCl and $K_2SO_4$ , respectively, are _____________.

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