Two elements A and B form compounds having formula $AB_2$ and $AB_4$. When dissolved in 20 g of benzene ($C_6H_6$), 1 g of $AB_2$ lowers the freezing point by 2.3 K whereas 1.0 g of $AB_4$ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol$^{-1}$. Calculate atomic masses of A and B.

Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g

Exercises · 1.21

AB_2

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Concept understanding — Colligative Properties Association

Colligative Properties: The Intuition First

Imagine you're at a party. The room is full of people dancing — that's your solvent molecules, moving freely. Now, someone brings in a few heavy, slow-moving guests who just stand in one spot — those are your solute particles. They don't dance, they don't interact much, they just take up space.

What happens? The dancers now have less room to move. They bump into the standing guests more often. The whole atmosphere changes — the dancers can't move as freely, they can't escape the room as easily, and the overall "energy" of the party shifts.

That's the core idea of colligative properties. When you add a non-volatile solute (like salt) to a solvent (like water), the solute particles don't do anything special — they just exist in the solution. But their mere presence changes four measurable properties of the solvent:

Vapour pressure decreases
Boiling point increases
Freezing point decreases
Osmotic pressure increases

The key insight: these changes depend only on the number of solute particles, not on what kind of particles they are. One molecule of sugar and one ion of salt (if they don't dissociate) affect these properties identically — provided they're the same number of particles.

Note

This is why "colligative" comes from the Latin colligatus meaning "bound together" — the properties are bound to the quantity of solute, not its identity.

The Precise Statement

Colligative properties are properties of a solution that depend solely on the ratio of the number of solute particles to the number of solvent molecules in a given solution, and not on the chemical nature of the solute.

Mathematically, for a dilute solution of a non-volatile, non-electrolyte solute:

\Delta P = P^0 \cdot x_2

Where:

$\Delta P$ = lowering of vapour pressure
$P^0$ = vapour pressure of pure solvent
$x_2$ = mole fraction of solute
$\Delta T_b$ = elevation in boiling point
$K_b$ = ebullioscopic constant (depends only on solvent)
$m$ = molality of solution
$\Delta T_f$ = depression in freezing point
$K_f$ = cryoscopic constant (depends only on solvent)
$\Pi$ = osmotic pressure
$i$ = van't Hoff factor (accounts for dissociation/association)
$M$ = molarity
$R$ = gas constant
$T$ = absolute temperature

The Crucial Distinction: Association vs. Dissociation

Now, here's where the association part comes in — and it's the twist that catches most students.

The formulas above assume the solute particles remain as individual, independent particles. But in reality:

Dissociation: Some solutes break apart into smaller particles (e.g., NaCl → Na⁺ + Cl⁻). This increases the number of particles, so the colligative effect is larger than expected.
Association: Some solutes clump together into larger particles (e.g., acetic acid in benzene forms dimers: 2 CH₃COOH → (CH₃COOH)₂). This decreases the number of particles, so the colligative effect is smaller than expected.

Watch out

A common mistake: students think "association" means the solute interacts with the solvent. No — association means solute particles bind to each other, reducing the effective particle count. Solvent-solute interactions affect non-colligative properties like solubility.

The van't Hoff Factor $i$

To account for these real-world effects, we introduce the van't Hoff factor:

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

For a non-electrolyte that doesn't associate or dissociate: $i = 1$

For dissociation (e.g., NaCl): $i > 1$ (ideally 2 for NaCl)

For association (e.g., acetic acid dimerizing): $i < 1$

The corrected formulas become:

\Delta T_b = i \cdot K_b \cdot m

\Delta T_f = i \cdot K_f \cdot m

\Pi = i \cdot MRT

A Concrete Example

Consider acetic acid (CH₃COOH) dissolved in benzene. In benzene, acetic acid molecules form hydrogen-bonded dimers:

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

If you dissolve 1 mole of acetic acid, you might end up with only 0.6 moles of particles (0.4 moles of dimers + 0.2 moles of monomers). So $i = 0.6$ . …

Why this formula?

Colligative Properties & Association: Why the Formula Holds

Let's build this from first principles — understanding why association changes colligative properties, not just memorizing the formula.

The Core Idea: What Are Colligative Properties?

Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. The four key ones are:

Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure

When a solute associates (e.g., two molecules dimerize), the effective number of particles decreases. This is the entire reason the formula changes.

The van't Hoff Factor: The Bridge

We define the van't Hoff factor $i$ as:

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

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

For association, $i < 1$ .

Example: Dimerization of Benzoic Acid in Benzene

Benzoic acid ( $C_6H_5COOH$ ) forms dimers in benzene:

2 \, C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2

If we dissolve $n$ moles of monomer, but only $n/2$ moles of dimer exist, then:

i = \frac{n/2}{n} = 0.5

Deriving the Modified Formula

Step 1: Start with the Normal Colligative Formula

For freezing point depression (the most common exam case):

\Delta T_f = K_f \cdot m

where $m$ is the molality of the solute (moles per kg solvent).

Step 2: Replace $m$ with Effective Molality

Because only the number of particles matters, we replace $m$ with $i \cdot m$ :

\Delta T_f = K_f \cdot (i \cdot m)

This is the general formula for any colligative property when association or dissociation occurs.

Step 3: Express $i$ in Terms of Degree of Association

Let:

$\alpha$ = degree of association (fraction of molecules that associate)
$n$ = number of molecules that combine to form one associated particle (e.g., $n=2$ for dimerization)

For a dimerization ( $n=2$ ):

Initially: 1 mole of monomer
After association: $(1 - \alpha)$ moles remain as monomer, and $\alpha/2$ moles of dimer form
Total particles = $(1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$

Thus:

i = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}

General formula for association of $n$ molecules:

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

Why This Makes Physical Sense …

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From $\Delta T_f = K_f\, m$ , the molar masses are $M_{AB_2} = 110.9$ and $M_{AB_4} = 196.2\ \text{g mol}^{-1}$ . Solving $a + 2b$ and $a + 4b$ gives atomic masses $A \approx 25.59$ u and $B \approx 42.64$ u.

Let $a$ and $b$ be the atomic masses of A and B. With $M = \dfrac{1000\, K_f\, w}{\Delta T_f\, W_{solvent}}$ , using $w = 1$ g, $W_{solvent} = 20$ g, $K_f = 5.1$ :

1. Molar mass of $AB_2$ ( $\Delta T_f = 2.3$ K):

M_{AB_2} = \frac{1000 \times 5.1 \times 1}{2.3 \times 20} = \frac{5100}{46} = 110.87\ \text{g mol}^{-1}

\Rightarrow\ a + 2b = 110.87 \quad (1)

2. Molar mass of $AB_4$ ( $\Delta T_f = 1.3$ K): …

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Method: Freezing Point Depression (Colligative Property)

This method uses the relationship between the lowering of freezing point ( $\Delta T_f$ ) and the molality of the solution to find molar masses, then deducing atomic masses from the formulas.

Step 1: Recall the formula

For a non-electrolyte solute:

\Delta T_f = K_f \cdot m

where:

$\Delta T_f$ = depression in freezing point (K)
$K_f$ = molal depression constant (K kg mol $^{-1}$ )
$m$ = molality = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Step 2: Write the expression for molar mass

Let $M$ = molar mass of solute (g mol $^{-1}$ ).

Moles of solute = $\frac{\text{mass of solute}}{M}$

Molality $m = \frac{\text{mass of solute}}{M \times \text{mass of solvent (kg)}}$

So:

\Delta T_f = K_f \cdot \frac{\text{mass of solute}}{M \times \text{mass of solvent (kg)}}

Rearranging:

M = \frac{K_f \times \text{mass of solute}}{\Delta T_f \times \text{mass of solvent (kg)}}

Step 3: Apply to $AB_2$

Given:

$K_f = 5.1$ K kg mol $^{-1}$
Mass of solute ( $AB_2$ ) = 1.0 g
Mass of solvent = 20 g = $0.020$ kg
$\Delta T_f = 2.3$ K

M_{AB_2} = \frac{5.1 \times 1.0}{2.3 \times 0.020}

M_{AB_2} = \frac{5.1}{0.046} = 110.87 \text{ g mol}^{-1}

Let atomic mass of A = $a$ , B = $b$ .

For $AB_2$ :

a + 2b = 110.87 \quad \text{(Equation 1)}

Step 4: Apply to $AB_4$

Given: …

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Concept understanding — Colligative Properties Association

Colligative Properties: The Intuition First

The Precise Statement

The Crucial Distinction: Association vs. Dissociation

The van't Hoff Factor $i$

A Concrete Example

Why this formula?

Colligative Properties & Association: Why the Formula Holds

The Core Idea: What Are Colligative Properties?

The van't Hoff Factor: The Bridge

Example: Dimerization of Benzoic Acid in Benzene

Deriving the Modified Formula

Step 1: Start with the Normal Colligative Formula

Step 2: Replace $m$ with Effective Molality

Step 3: Express $i$ in Terms of Degree of Association

For a dimerization ( $n=2$ ):

General formula for association of $n$ molecules:

Why This Makes Physical Sense …

Method: Freezing Point Depression (Colligative Property)

Step 1: Recall the formula

Step 2: Write the expression for molar mass

Step 3: Apply to $AB_2$

Step 4: Apply to $AB_4$

Mistake 1: Inventing a Van't Hoff Factor That Isn't There

Mistake 2: Using Mass of Solvent in Grams Without Converting to kg

Mistake 3: Not Setting Up Two Independent Equations

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