Which of the following aqueous solutions should have the highest boiling point? (i) 1.0MNaOH (ii) 1.0MNa₂SO₄ (

NCERT Exemplar · Q9

Q. Which of the following aqueous solutions should have the highest boiling point?

(i)

1.0\ M\ NaOH

(ii)

1.0\ M\ Na_2SO_4

(iii)

1.0\ M\ NH_4NO_3

(iv)

1.0\ M\ KNO_3

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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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Boiling point elevation depends on the total number of particles in solution. $\text{Na}_2\text{SO}_4$ dissociates into three ions per formula unit, giving the highest van't Hoff factor and thus the highest boiling point.

The boiling point of a solution rises above that of the pure solvent because solute particles disrupt the solvent's ability to escape into the vapor phase - a colligative property depending only on the number of dissolved particles.

\Delta T_b = i \cdot K_b \cdot m

Since all solutions here have the same concentration (1.0 M, approximately the same molality for dilute aqueous solutions), the solution with the largest $i$ has the highest boiling point.

NaOH: $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$ , $i=2$ . …

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Concept: Colligative Properties — Elevation in Boiling Point

The boiling point elevation depends on the number of solute particles in solution, not the type of particle. The formula is:

\Delta T_b = i \cdot K_b \cdot m

Where:

$\Delta T_b$ = boiling point elevation
$i$ = van’t Hoff factor (number of ions per formula unit)
$K_b$ = ebullioscopic constant (same for water here)
$m$ = molality (same for all options — 1.0 M is approximately 1.0 m for dilute aqueous solutions)

Since $K_b$ and $m$ are the same, the solution with the largest $i$ will have the highest boiling point.

Method: Van’t Hoff Factor Comparison

Steps:

Write the dissociation equation for each solute in water.
Count the total number of ions produced per formula unit — this is $i$ .
Compare $i$ values — larger $i$ → higher boiling point.

Applying the steps:

Solute	Dissociation	$i$ (ions)
NaOH	$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$	2

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Common Mistakes & How to Avoid Them

Mistake 1: Assuming all 1.0 M solutions have the same boiling point

Why students make it: They see "1.0 M" for all options and think concentration alone determines boiling point.

How to avoid: Remember that boiling point elevation depends on the total number of particles in solution, not just the molarity of the solute. The formula is:

\Delta T_b = i \cdot K_b \cdot m

where $i$ = van't Hoff factor (number of particles per formula unit), $K_b$ = ebullioscopic constant, and $m$ = molality.

Key insight: For aqueous solutions at the same molarity, the solute that dissociates into the most ions gives the highest boiling point.

Mistake 2: Forgetting to count ALL ions from dissociation

Why students make it: They count only the obvious ions and miss one.

How to avoid: Write the dissociation equation for each compound:

Compound	Dissociation	Total ions ( $i$ )
NaOH	$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$	2
Na $_2$ SO $_4$	$\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}$	3
NH $_4$ NO $_3$	$\text{NH}_4\text{NO}_3 \rightarrow \text{NH}_4^+ + \text{NO}_3^-$	2
KNO $_3$	$\text{KNO}_3 \rightarrow \text{K}^+ + \text{NO}_3^-$	2

Correct answer: Option (ii) 1.0 M Na $_2$ SO $_4$ has the highest boiling point because it produces 3 ions per formula unit.

Mistake 3: Confusing molarity with molality

Why students make it: The problem gives concentrations in molarity (M), but the boiling point formula uses molality (m).

How to avoid: For dilute aqueous solutions (like 1.0 M), molarity and molality are approximately equal because 1 L of water ≈ 1 kg of water. In exam problems, treat them as equal unless told otherwise. …

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Q.Which of the following aqueous solutions should have the highest boiling point?

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