An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at 27°C was f

Question

Q. An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at 27°C was found to be 25 g mol⁻¹. Calculate the percentage dissociation of NaOH in this solution. [Atomic mass : Na = 23 u, O = 16 u, H = 1 u]

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

The observed molar mass (25 g/mol) is lower than the theoretical molar mass (40 g/mol) because NaOH dissociates into ions. Using the van’t Hoff factor $i = \frac{\text{theoretical molar mass}}{\text{observed molar mass}} = 1.6$ , and relating $i$ to the degree of dissociation $\alpha$ for a binary electrolyte ( $i = 1 + \alpha$ ), we find $\alpha = 0.6$ , i.e., 60% dissociation.

The key here is that osmotic pressure is a colligative property — it depends only on the number of solute particles in solution, not on their identity. When NaOH dissolves, it dissociates into Na⁺ and OH⁻ ions. This increases the total number of particles, which in turn increases the osmotic pressure. If we measure osmotic pressure and use it to calculate molar mass, we get an apparent (observed) molar mass that is lower than the true molar mass. The ratio of the true molar mass to the observed molar mass gives the van’t Hoff factor $i$ , which tells us how many particles each formula unit produces on average.

Let’s walk through the calculation.

Find the theoretical molar mass of NaOH.

Atomic masses: Na = 23, O = 16, H = 1.

So $M_{\text{theoretical}} = 23 + 16 + 1 = 40 \text{ g mol}^{-1}$ .
Understand the relationship between observed molar mass and the van’t Hoff factor.

For a colligative property like osmotic pressure ( $\Pi = iCRT$ ), the observed molar mass $M_{\text{obs}}$ is related to the true molar mass $M_{\text{theo}}$ by:

i = \frac{M_{\text{theo}}}{M_{\text{obs}}}

This is because if there were no dissociation, $i = 1$ and $M_{\text{obs}} = M_{\text{theo}}$ . With dissociation, more particles mean higher $\Pi$ , so the calculated $M_{\text{obs}}$ is smaller.

Plug in the given values. $M_{\text{obs}} = 25 \text{ g mol}^{-1}$ , $M_{\text{theo}} = 40 \text{ g mol}^{-1}$ .

i = \frac{40}{25} = 1.6

Relate $i$ to the degree of dissociation $\alpha$ for NaOH.

NaOH dissociates as: $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$ .

Initially, we have 1 mole of NaOH. After dissociation, if $\alpha$ is the fraction dissociated:
- Moles of NaOH remaining = $1 - \alpha$
- Moles of Na⁺ formed = $\alpha$
- Moles of OH⁻ formed = $\alpha$ Total moles after dissociation = $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$ .
The van’t Hoff factor $i$ is defined as: …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: All ______ properties of solution related with decreasing in vapour pressure.$
›Reveal solutionSolution

The blank is colligative.

Colligative properties are those that depend only on the number of solute particles present, not on their nature. All four of them - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure - are consequences of th …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Explain the following — Colligative properties$
›Reveal solutionSolution
Colligative properties depend on solute particle count relative to solvent, not on what the solute chemically is.

Colligative properties are those properties of a solution that depend only on the number (concentration) of solute particles present in the solution, and not on the nature/identity of the solute particles. The four colligative properties are:

Relative lowering of vapour pressure

Elevation of boiling point

Depression of freezing point …
Start your 14-day free trial to see the full solution →
CBSE 2021Set ANNUAL1 markMCQ
$Q. The colligative properties of a dilute solution depend on -$
(a) nature of the solute
(b) nature of the solvent
(c) number of solute particles
(d) molecular mass of solute
›Reveal solutionSolution

Colligative properties depend only on how many solute particles are dissolved, not on what they are.

Colligative properties (relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, osmotic pressure) are, by definition, properties that depend solely on the number of solute particles present in a given amount of solvent — not on the chemical identity, size, or molecular mass of the solute. This is why a dissociating electrolyte …

Start your 14-day free trial to see the full solution →

Q.An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at 27°C was found to be 25 g mol⁻¹. Calculate the percentage dissociation of NaOH in this solution. [Atomic mass : Na = 23 u, O = 16 u, H = 1 u]

🎓Unlock everything free for 14 days

Unlock everything free for 14 days