Q.When 19.5 g of $F-CH_2-COOH$ (Molar mass = 78 g mol$^{-1}$), is dissolved in 500 g of water, the depression in freezing point is observed to be 1 °C. Calculate the degree of dissociation of $F-CH_2-COOH$. [Given : $K_f$ for water = 1.86 K kg mol$^{-1}$]

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:

1. Vapour pressure decreases
2. Boiling point increases
3. Freezing point decreases
4. 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.

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:

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.

The van't Hoff Factor $i$

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

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:

A Concrete Example

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

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 $i = 0.6$. …