In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M $MgCl_2$ solution is _____________. (i) the same (ii) about twice (iii) about three times (iv) about six times

> [!TLDR] > The depression in freezing point depends on the number of solute particles. $MgCl_2$ dissociates into three ions, while glucose does not dissociate, leading to about three times the depression in freezing point for $MgCl_2$ compared to glucose at the same concentration. Colligative properties are fascinating because they depend solely on the *number* of solute particles in a solution, not on their chemical identity. Depression in freezing point is one such property. When a solute is added to a solvent, it interferes with the solvent's ability to form a crystal lattice, thus lowering the freezing point. The more particles present, the greater this interference, and the larger the depression in freezing point. The Van't Hoff factor, denoted by $i$, is crucial here. It accounts for the effective number of particles produced when a solute dissolves. * For non-electrolytes (like glucose), which do not dissociate into ions, $i=1$. One molecule dissolved yields one particle. * For electrolytes (like $MgCl_2$), which dissociate into ions, $i$ is approximately equal to the number of ions produced per formula unit, assuming complete dissociation. Let's apply this understanding to the given problem. 1. **Recall the formula for Depression in Freezing Point.** The depression in freezing point ($\Delta T_f$) is directly proportional to the molality ($m$) of the solution and the Van't Hoff factor ($i$). > [!FORMULA] > $\Delta T_f = i K_f m$ > where $K_f$ is the cryoscopic constant (molal depression constant) of the solvent. For a given solvent (water, in this case), $K_f$ is constant. Both solutions are 0.01 M. For dilute aqueous solutions, molarity (M) is a good approximation for molality (m) because the density of water is close to $1 \text{ g/mL}$, meaning $1 \text{ L}$ of solution is approximately $1 \text{ kg}$ of solvent. Therefore, we can consider the molality ($m$) to be the same for both solutions. Since $K_f$ and $m$ are the

In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl₂ solution is

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MgCl_2

(i) the same

(ii) about twice

(iii) about three times

(iv) about six times

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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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The depression in freezing point depends on the number of solute particles. $MgCl_2$ dissociates into three ions, while glucose does not dissociate, leading to about three times the depression in freezing point for $MgCl_2$ compared to glucose at the same concentration.

Colligative properties are fascinating because they depend solely on the number of solute particles in a solution, not on their chemical identity. Depression in freezing point is one such property. When a solute is added to a solvent, it interferes with the solvent's ability to form a crystal lattice, thus lowering the freezing point. The more particles present, the greater this interference, and the larger the depression in freezing point.

The Van't Hoff factor, denoted by $i$ , is crucial here. It accounts for the effective number of particles produced when a solute dissolves.

For non-electrolytes (like glucose), which do not dissociate into ions, $i=1$ . One molecule dissolved yields one particle.
For electrolytes (like $MgCl_2$ ), which dissociate into ions, $i$ is approximately equal to the number of ions produced per formula unit, assuming complete dissociation.

Let's apply this understanding to the given problem.

Recall the formula for Depression in Freezing Point.

The depression in freezing point ( $\Delta T_f$ ) is directly proportional to the molality ( $m$ ) of the solution and the Van't Hoff factor ( $i$ ).
$\Delta T_f = i K_f m$
Both solutions are 0.01 M. For dilute aqueous solutions, molarity (M) is a good approximation for molality (m) because the density of water is close to $1 \text{ g/mL}$ , meaning $1 \text{ L}$ of solution is approximately $1 \text{ kg}$ of solvent. Therefore, we can consider the molality ( $m$ ) to be the same for both solutions.

Since $K_f$ and $m$ are the same for both solutions, the ratio of their freezing point depressions will simply be the ratio of their Van't Hoff factors.
Determine the Van't Hoff factor for Glucose.

Glucose ( $C_6H_{12}O_6$ ) is a non-electrolyte. When dissolved in water, it does not dissociate into ions. Each glucose molecule remains intact.

Therefore, for glucose, the Van't Hoff factor $i_{glucose} = 1$ .

The depression in freezing point for the glucose solution is:

$\Delta T_{f, glucose} = 1 \times K_f \times 0.01$
Determine the Van't Hoff factor for $MgCl_2$ .

Magnesium chloride ( $MgCl_2$ ) is an ionic compound and a strong electrolyte. When dissolved in water, it dissociates completely into its constituent ions.

The dissociation reaction is: …

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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.In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M $MgCl_2$ solution is _____________.

Concept understanding — VanT Hoff Factor Association

The Intuition: What Happens When Particles Stick Together?

The Van't Hoff Factor: The Correction Number

A Concrete Example: Acetic Acid in Benzene

The General Formula for Association

How Association Affects Colligative Properties

Why this formula?

Van't Hoff Factor for Association: Why the Formula Holds

1. The Core Idea: What Changes?

2. Setting Up the Association Process

What happens to the particles?

3. Total Particles After Association

4. The Van't Hoff Factor Formula

5. Why This Makes Physical Sense

Concept: Colligative Properties — Depression in Freezing Point

Method: Van’t Hoff Factor Comparison

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that $MgCl_2$ dissociates into ions

Mistake 2: Counting the wrong number of particles

Mistake 3: Confusing depression in freezing point with boiling point elevation

Why freezing point depends on particle count

Calculating the van't Hoff factor for each solute

Q.In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2MgCl_2MgCl2​ solution is _____________.

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Q.In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M $MgCl_2$ solution is _____________.

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