Why is the mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it wi

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Q. Why is the mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it with the help of Van't Hoff factor.

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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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When a solute associates or dissociates in solution, the number of particles changes, making the observed colligative property abnormal. The Van’t Hoff factor $i$ corrects for this, so the experimentally determined molar mass is either higher (association) or lower (dissociation) than the true molar mass.

The problem asks: why does the mass determined by measuring a colligative property sometimes come out abnormal? And how does the Van’t Hoff factor explain this?

Let’s start with the core idea. Colligative properties — like freezing point depression, boiling point elevation, and osmotic pressure — depend only on the number of solute particles in solution, not on their identity. When you dissolve a substance, you expect a certain number of particles based on its formula mass. But some solutes behave differently.

For example, sodium chloride ( $\text{NaCl}$ ) in water splits into $\text{Na}^+$ and $\text{Cl}^-$ ions. One formula unit gives two particles. So the actual number of particles is more than expected. Conversely, benzoic acid in benzene forms dimers — two molecules stick together, so the number of particles is less than expected.

Because colligative properties are proportional to particle count, an abnormal particle count gives an abnormal reading. If you then use that reading to calculate molar mass (using the usual formulas), you get a value that is not the true molar mass — it’s an apparent or abnormal molar mass.

The Van’t Hoff factor $i$ is the tool that quantifies this deviation.

i = \frac{\text{observed number of particles}}{\text{expected number of particles (if no association/dissociation)}}

For a non-electrolyte that neither associates nor dissociates, $i = 1$ . For dissociation, $i > 1$ ; for association, $i < 1$ .

Now, the relationship between observed molar mass ( $M_{\text{obs}}$ ) and true molar mass ( $M_{\text{true}}$ ) is:

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

Why? Because colligative properties are inversely proportional to molar mass. If the observed colligative effect is larger than expected (more particles), the calculated molar mass comes out smaller. So $M_{\text{obs}} < M_{\text{true}}$ , and $i > 1$ . If the effect is smaller (fewer particles), $M_{\text{obs}} > M_{\text{true}}$ , and $i < 1$ .

Let’s walk through the reasoning step by step.

Recall the basic colligative formula. For freezing point depression, $\Delta T_f = K_f \cdot m$ , where $m$ is molality. Molality is moles of solute per kg of solvent. If you know $\Delta T_f$ and $K_f$ , you can calculate $m$ , and from $m$ and the mass of solute used, you get the molar mass: $M = \frac{\text{mass of solute}}{m \times \text{kg solvent}}$ .
Now introduce the abnormal behaviour. Suppose the solute dissociates. The actual number of particles in solution is greater than the number of formula units dissolved. So the observed $\Delta T_f$ is larger than expected for the given mass of solute. Plugging this larger $\Delta T_f$ into the formula gives a larger $m$ , and therefore a smaller calculated molar mass — $M_{\text{obs}}$ is less than $M_{\text{true}}$ .
For association, the opposite happens. Fewer particles mean a smaller $\Delta T_f$ , a smaller $m$ , and a larger calculated molar mass — $M_{\text{obs}}$ is greater than $M_{\text{true}}$ .
The Van’t Hoff factor corrects this. The true colligative property is related to the observed one by:

\Delta T_f (\text{observed}) = i \cdot \Delta T_f (\text{expected})

Since $\Delta T_f \propto \frac{1}{M}$ , we get:

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

Watch out

A common mistake is to think $i = M_{\text{obs}} / M_{\text{true}}$ . Check: if dissociation occurs, $M_{\text{obs}}$ is smaller, so $i$ should be greater than 1. The correct relation is $i = M_{\text{true}} / M_{\text{obs}}$ , which gives $i > 1$ when $M_{\text{obs}} < M_{\text{true}}$ . …

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Van't Hoff Factor & Abnormal Molar Masses (Association)

Why is the mass "abnormal"?

When we measure a colligative property (like freezing point depression or osmotic pressure) to find the molar mass of a solute, we assume the solute particles behave independently in solution.

However, for some solutes (like benzoic acid in benzene or acetic acid in benzene), the molecules associate (stick together) to form dimers or larger clusters. This means:

Fewer particles are present in solution than expected.
The colligative property (which depends on number of particles) is smaller than expected.
The calculated molar mass comes out higher than the actual molar mass.

This is called an abnormal molar mass.

Method: Van't Hoff Factor ( $i$ ) for Association

Name of method: Van't Hoff Factor correction for association.

Concept: The Van't Hoff factor $i$ is defined as:

i = \frac{\text{Observed colligative property}}{\text{Expected colligative property}} = \frac{\text{Normal molar mass}}{\text{Observed molar mass}}

For association, $i < 1$ .

Steps to solve a typical problem

Step 1: Write the association equilibrium

For example, if two molecules of solute $A$ associate to form a dimer $A_2$ :

2A \rightleftharpoons A_2

Step 2: Define degree of association ( $\alpha$ )

Let $\alpha$ = fraction of $A$ that associates.

Initial moles of $A$ = 1 (or $n$ )
Moles of $A$ that associate = $\alpha$
Moles of $A$ left unassociated = $1 - \alpha$
Moles of $A_2$ formed = $\frac{\alpha}{2}$ (since 2 molecules make 1 dimer)

Step 3: Calculate total number of particles after association

\text{Total moles} = (1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}

Step 4: Apply Van't Hoff factor

i = \frac{\text{Observed number of particles}}{\text{Expected number of particles}} = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}

Step 5: Relate $i$ to molar masses …

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Here’s a breakdown of the common mistakes students make on this topic, along with clear strategies to avoid them.

1. Confusing “Abnormal Mass” with “Wrong Experiment”

The Mistake:

Students often think “abnormal” means the lab experiment failed or the balance was faulty. They miss the core idea: the mass appears abnormal because the number of particles in solution is different from what we assumed.

Why it’s wrong:

Colligative properties depend only on the number of solute particles, not their identity. If a solute associates (forms dimers, trimers) or dissociates (breaks into ions), the actual particle count changes — so the calculated molar mass becomes “abnormal.”

How to Avoid:

Always ask: “Does this solute stay as single molecules in solution?”
Remember: Abnormal mass = calculated mass using colligative property ≠ theoretical molar mass because the particle count is different.

2. Forgetting the Van’t Hoff Factor Definition

The Mistake:

Students write $i = \frac{\text{Observed colligative property}}{\text{Expected colligative property}}$ but then plug in masses instead of particle numbers.

Why it’s wrong:

The Van’t Hoff factor $i$ is defined as:

i = \frac{\text{Actual number of particles in solution}}{\text{Number of particles if no association/dissociation}}

It directly links to molar mass:

i = \frac{\text{Normal molar mass}}{\text{Observed (abnormal) molar mass}}

How to Avoid:

Memorise the two equivalent forms of $i$ $i$ :
- For colligative property: $i = \frac{\text{Observed } \Delta T_f}{\text{Expected } \Delta T_f}$
- For molar mass: $i = \frac{M_{\text{normal}}}{M_{\text{observed}}}$
Practice converting between them.

3. Mixing Up Association vs. Dissociation

The Mistake:

Students treat association (e.g., benzoic acid dimerising in benzene) the same as dissociation (e.g., NaCl splitting into ions). They use the same formula for $i$ without adjusting for the number of particles formed.

Why it’s wrong:

Association → particles decrease → $i < 1$ → observed molar mass increases (appears heavier).
Dissociation → particles increase → $i > 1$ → observed molar mass decreases (appears lighter).

How to Avoid:

Draw a simple particle diagram before calculating.
For association: if $n$ molecules combine, $i = \frac{1}{n}$ (for complete association).
For dissociation: if one molecule gives $n$ ions, $i = n$ (for complete dissociation).

4. Using the Wrong Formula for Degree of Association/Dissociation

The Mistake:

Students directly write $i = 1 + (n-1)\alpha$ for dissociation but then use the same for association without changing the sign.

Why it’s wrong:

The correct formulas are:

For dissociation:

i = 1 + (n-1)\alpha

where $\alpha$ = degree of dissociation, $n$ = number of ions.

For association:

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

where $\alpha$ = degree of association, $n$ = number of molecules that associate.

How to Avoid:

Write the chemical equation first (e.g., $nA \rightleftharpoons A_n$ ).
Count initial moles and equilibrium moles. Derive $i$ from the ratio — don’t memorise blindly.

5. Ignoring the Solvent’s Role

The Mistake:

Students assume association/dissociation happens the same way in every solvent. For example, they treat acetic acid in water (dissociates) the same as in benzene (associates).

Why it’s wrong:

Polar solvents (water) favour dissociation (ions stabilised).
Non-polar solvents (benzene) favour association (hydrogen bonding between solute molecules).

How to Avoid: …

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

CBSE 2026Set 56/1/11 markMCQ
$CaCl_2$
›Reveal solutionSolution
Freezing point depression depends on the number of particles in solution, not just the solute concentration. The solution with the most ions per formula unit will have the lowest freezing point. Here, 0.1 M $CaCl_2$ gives 3 particles per formula unit, the highest among the options, so it has the lowest freezing point.

The key idea is colligative properties — properties that depend only on the number of solute particles, not on their identity. Freezing point depression is one such property. The more particles you have in solution, the more the freezing point drops.

For ionic compounds, each formula unit dissociates into multiple ions. So a 0.1 M solution of $CaCl_2$ doesn't just give 0.1 moles of particles per litre — it gives more, because each $CaCl_2$ breaks into one $Ca^{2+}$ and two $Cl^-$ ions. That's three particles total. Compare that to glucose or urea, which are covalent and don't dissociate at all — they give just one particle per molecule.

The van't Hoff factor $i$ captures this: it's the actual number of particles per formula unit in solution. For non-electrolytes like glucose and urea, $i = 1$ . For $KCl$ , which dissociates into $K^+$ and $Cl^-$ , $i = 2$ . For $CaCl_2$ , $i = 3$ (assuming complete dissociation).
$\Delta T_f = i \cdot K_f \cdot m$
Since $K_f$ and $m$ are the same for all options (all 0.1 M in water), the freezing point depression is directly proportional to $i$ . The larger $i$ is, the lower the freezing point.

Let's check each option:

0.1 M Glucose — Glucose is a covalent molecule. It does not dissociate. So $i = 1$ . Freezing point depression is $\Delta T_f = 1 \cdot K_f \cdot 0.1$ .

0.1 M Urea — Urea is also covalent and non-electrolytic. $i = 1$ . Same depression as glucose.

0.1 M $KCl$ — $KCl$ dissociates completely: $KCl \rightarrow K^+ + Cl^-$ . That's 2 ions. So $i = 2$ . Depression is $\Delta T_f = 2 \cdot K_f \cdot 0.1$ — twice that of glucose or urea. …
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CBSE 2025Set 56/5/11 markMCQ
$Na_2SO_4$
›Reveal solutionSolution
The solution with the smallest van't Hoff factor produces the fewest particles and thus the highest freezing point. Glucose (a non-electrolyte) gives $i = 1$ , while all ionic compounds dissociate into multiple ions. The answer is (C) 1·0 M Glucose.

Why freezing point depends on particle count

Freezing point depression is a colligative property—it depends only on the number of solute particles, not their identity. The relationship is:
$\Delta T_f = i \cdot K_f \cdot m$
where $i$ is the van't Hoff factor (the number of particles each formula unit produces in solution), $K_f$ is the cryoscopic constant, and $m$ is molality.

Since all solutions here have the same concentration (1.0 M, approximately 1.0 m for dilute aqueous solutions) and the same solvent (water, so same $K_f$ ), the depression depends entirely on $i$ . The solution with the smallest $i$ experiences the least depression and therefore has the highest freezing point.

Calculating the van't Hoff factor for each solute

Let's determine how many particles each compound produces when it dissolves:

KCl (potassium chloride) This strong electrolyte dissociates completely:

$\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$
Each formula unit produces 2 ions, so $i = 2$ .

$\text{Na}_2\text{SO}_4$ (sodium sulfate) Complete dissociation gives:

$\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}$
Each formula unit produces 3 ions, so $i = 3$ .

Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) Glucose is a non-electrolyte—it dissolves as intact molecules without dissociating:

$\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_6\text{H}_{12}\text{O}_6$
Each molecule remains one particle, so $i = 1$ .

$\text{AlCl}_3$ (aluminum chloride) Complete dissociation yields:

$\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-$
Each formula unit produces 4 ions, so $i = 4$ . …
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CBSE 2025Set 56/6/11 markMCQ
$K_f$
›Reveal solutionSolution
For a completely dissociated 1 molal KCl solution, the van’t Hoff factor $i = 2$ . The freezing point depression is $\Delta T_f = i \cdot K_f \cdot m = 2 \times 1.86 \times 1 = 3.72\ \text{K}$ , so the freezing point is $0 - 3.72 = -3.72\ ^\circ\text{C}$ . The correct option is (A).

When a non-volatile solute dissolves in a solvent, the freezing point of the solution is lower than that of the pure solvent. The key idea is that the depression depends on the total number of particles in solution, not just the number of formula units dissolved. For an electrolyte like KCl, which dissociates completely into K⁺ and Cl⁻, each mole of KCl gives two moles of ions. The van’t Hoff factor $i$ captures this: it is the ratio of the actual number of particles after dissociation to the number of formula units dissolved.

For KCl, complete dissociation means $i = 2$ .

The formula for freezing point depression is:
$\Delta T_f = i \cdot K_f \cdot m$
where $K_f$ is the cryoscopic constant (here $1.86\ \text{K kg mol}^{-1}$ ) and $m$ is the molality (here $1\ \text{mol kg}^{-1}$ ).

Now let’s work through it step by step.

Identify the van’t Hoff factor.

KCl dissociates as: $\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$ . One formula unit yields two ions. Since the problem states “completely dissociated”, $i = 2$ .

Plug into the depression formula.

$\Delta T_f = 2 \times 1.86\ \text{K kg mol}^{-1} \times 1\ \text{mol kg}^{-1} = 3.72\ \text{K}$
Because the freezing point constant $K_f$ is given in Kelvin, the depression $\Delta T_f$ is also in Kelvin. But since a change of 1 K equals a change of 1 °C, we can directly say $\Delta T_f = 3.72\ ^\circ\text{C}$ .

Apply the depression to the pure solvent’s freezing point. …
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CBSE 2024Set 56/1/11 markMCQ
$Q. For an electrolyte undergoing association in a solvent, the van't Hoff factor: (A) is always greater than one (B) has negative value (C) has zero value (D) is always less than one$
›Reveal solutionSolution
When solute particles associate in a solvent, the total number of independent particles decreases, leading to a van't Hoff factor that is always less than one.

The van't Hoff factor, denoted by $i$ , is a crucial concept in understanding the behavior of solutions, especially when dealing with colligative properties. Colligative properties (like elevation in boiling point, depression in freezing point, osmotic pressure, and relative lowering of vapor pressure) depend solely on the number of solute particles in a solution, not on their identity or nature.

However, this ideal behavior is observed only for non-electrolytes that do not undergo any change in the solution. When an electrolyte is dissolved, it can either dissociate (break into more particles) or associate (combine to form fewer, larger particles). The van't Hoff factor accounts for these deviations from ideal behavior.

Conceptually, the van't Hoff factor is the ratio of the observed colligative property to the theoretical (calculated assuming no association or dissociation) colligative property. More fundamentally, it represents the ratio of the actual number of moles of particles in solution after association or dissociation to the number of moles of solute initially dissolved.
$i = \frac{\text{Observed Colligative Property}}{\text{Normal (Theoretical) Colligative Property}}$
When an electrolyte undergoes association, it means that multiple solute particles combine to form a single, larger aggregate. For example, two acetic acid molecules can associate to form a dimer in benzene. This process reduces the total number of independent particles in the solution. Since colligative properties depend on the number of particles, a reduction in particles will lead to a smaller observed colligative property compared to what would be expected if no association occurred.

Let's walk through the reasoning step-by-step:

Understanding Association:

Consider a solute 'A' that associates in a solvent. If $n$ molecules of 'A' combine to form one associated molecule 'A $_n$ ', the process can be represented as:

$n\text{A} \rightleftharpoons \text{A}_n$

This means that $n$ individual particles effectively become one particle.

Effect on Number of Particles:

Suppose we initially dissolve 1 mole of solute 'A'. If a fraction $\alpha$ of these molecules associate, then:

Moles of 'A' remaining unassociated $= (1 - \alpha)$ moles.

Moles of 'A $_n$ ' formed from the associated fraction $= \frac{\alpha}{n}$ moles (because $n$ moles of 'A' form 1 mole of 'A $_n$ ').

The total number of moles of particles in the solution after association is the sum of unassociated 'A' and associated 'A $_n$ ':

Total moles of particles $= (1 - \alpha) + \frac{\alpha}{n}$

Calculating the van't Hoff Factor ( $i$ ):

Using the definition of $i$ as the ratio of actual moles of particles to initial moles of solute:

$i = \frac{\text{Total moles of particles after association}}{\text{Initial moles of solute}}$

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

$i = 1 - \alpha + \frac{\alpha}{n}$

Analyzing the Value of $i$ for Association: …
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CBSE 2024Set 56/1/11 markMCQ
$Q. Assertion (A): When NaCl is added to water a depression in freezing point is observed. Reason (R): NaCl undergoes dissociation in water. [Select the correct code: (A) Both A and R are true and R is the correct explanation of A; (B) Both A and R are true but R is not the correct explanation of A; (C) A is true but R is false; (D) A is false but R is true.]$
›Reveal solutionSolution
Adding NaCl to water causes freezing point depression because NaCl dissociates into ions, increasing the number of solute particles, which is the fundamental basis of colligative properties. Both the assertion and reason are true, and the reason correctly explains the assertion.

When we talk about properties of solutions, some of them depend only on the number of solute particles present, not on their specific chemical identity. These are called colligative properties. Freezing point depression is one such property.

The core idea behind freezing point depression is that the presence of solute particles interferes with the solvent molecules' ability to arrange themselves into a stable solid crystal lattice. To overcome this interference and form the solid, a lower temperature is required. The more solute particles there are, the greater this interference, and thus the greater the depression in the freezing point.

For non-electrolytes (like sugar), one molecule of solute contributes one particle to the solution. However, for electrolytes (like NaCl), the situation is different. When an electrolyte dissolves in water, it dissociates into ions. This dissociation increases the effective number of particles in the solution. For example, one formula unit of NaCl breaks down into one $\text{Na}^+$ ion and one $\text{Cl}^-$ ion, effectively doubling the number of particles compared to a non-electrolyte of the same molar concentration.

This increase in the number of particles due to dissociation is quantified by the van 't Hoff factor ( $i$ ). For an ideal strong electrolyte like NaCl, which dissociates into two ions, $i$ is approximately 2. The formula for freezing point depression explicitly includes this factor:
$\Delta T_f = i K_f m$
Now, let's analyze the given assertion and reason.

Analyze Assertion (A): "When NaCl is added to water a depression in freezing point is observed."

Sodium chloride (NaCl) is a solute, and water is a solvent. Adding any non-volatile solute to a solvent will cause a depression in its freezing point because freezing point depression is a colligative property. Therefore, this assertion is true.

Analyze Reason (R): "NaCl undergoes dissociation in water."

NaCl is an ionic compound and a strong electrolyte. When dissolved in water, it completely dissociates into its constituent ions: $\text{Na}^+$ and $\text{Cl}^-$ .

The dissociation reaction is: $\text{NaCl}(aq) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq)$ .

Therefore, this reason is also true.

Determine if Reason (R) is the correct explanation for Assertion (A). …
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CBSE 2019Set ANNUAL1 mark
$Q. When the value of van't Hoff factor is less than one, this shows that the solute undergoes __________ in the solution.$
›Reveal solutionSolution

i < 1 indicates association of the solute.

The van't Hoff factor i = (observed number of particles)/(number of formula units dissolved). When solute molecules combine (associate) in solution — for example benzoic acid dimerising in benzene through hydrogen bonding — the number of particles b …

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CBSE 2018Set ANNUAL1 mark
$Q. Van't Hoff's factor for a solution is less than one, what is the conclusion drawn from it.$
›Reveal solutionSolution
$i<1$ tells us the solute is associated in the solution (e.g. dimerisation), giving fewer particles than the formula predicts.

Concept. The van't Hoff factor is defined as

$i = \frac{\text{observed (actual) number of particles in solution}}{\text{number of particles expected from the formula}}$

Colligative properties depend on the number of solute particles. $i$ corrects for dissociation or association.

Reasoning.

If $i>1$ : the solute dissociates (e.g. NaCl → Na $^+$ + Cl $^-$ ), giving more particles.

If $i=1$ : no change (normal solute).

If $i<1$ : the solute associates, so several molecules join to form fewer, larger particles (e.g. acetic acid or benzoic acid dimerises in benzene through hydrogen bonding). …
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Q.Why is the mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it with the help of Van't Hoff factor.

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