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

NCERT Exemplar · Q11

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.

Showing the 12 most recent of 17 on this concept.

COMEDK 2026Set 2026-M1 markMCQ
$1.04 \mathrm{~g} / \mathrm{ml}$
›Reveal solutionSolution

From $\Delta T_f=i\,K_f\,m$ the van't Hoff factor is $i\approx1.06$ , so dissociation $\alpha=i-1\approx0.06=6\%$ — option (A).

Moles of acetic acid ( $\text{CH}_3\text{COOH}$ , $M=60\ \text{g mol}^{-1}$ ):
$\text{mass}=0.8\ \text{mL}\times1.04\ \text{g mL}^{-1}=0.832\ \text{g},$ $n=\frac{0.832}{60}=0.013867\ \text{mol}.$
Molality (solvent water $=1.2\ \text{L}\approx1.2\ \text{kg}$ ):
$m=\frac{0.013867}{1.2}=0.011556\ \text{mol kg}^{-1}.$
Van't Hoff factor from the freezing-point depression $\Delta T_f=i\,K_f\,m$ :

$i=\frac{\Delta T_f}{K_f\,m}=\frac{0.0228}{1.86\times0.011556}=\frac{0.0228}{0.021494}=1.061.$ …
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COMEDK 2026Set 2026-M1 markMCQ
$Q. 0.02 M solution of sucrose is isotonic with 0.008 M solution of sodium sulphate. What is the percentage dissociation of the electrolyte? (A) 75 (B) 79 (C) 85 (D) 88$
›Reveal solutionSolution
Isotonic solutions have the same osmotic pressure. For sucrose (non‑electrolyte) and Na₂SO₄ (electrolyte), equate their van’t Hoff factors: $i_{\text{sucrose}} = 1$ and $i_{\text{Na₂SO₄}} = 1 + 2\alpha$ (where $\alpha$ is the degree of dissociation). Solving $0.02 \times 1 = 0.008 \times (1 + 2\alpha)$ gives $\alpha = 0.75$ or 75%. The correct option is (A).

Concept and Intuition

Two solutions are isotonic when they exert the same osmotic pressure across a semipermeable membrane. Osmotic pressure $\Pi$ is given by the van’t Hoff equation:
$\Pi = i \, C \, R T$
where $C$ is the molar concentration, $R$ the gas constant, $T$ the temperature, and $i$ the van’t Hoff factor — the number of particles per formula unit in solution.

Sucrose is a non‑electrolyte; it does not dissociate. So $i = 1$ .

Sodium sulphate ( $\text{Na}_2\text{SO}_4$ ) dissociates in water:

$\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}$
If dissociation is complete, $i = 3$ . But here it is partial — we are asked for the percentage dissociation.

Because both solutions are at the same temperature, the isotonic condition simplifies to:
$i_1 C_1 = i_2 C_2$
We know $C_1 = 0.02\,\text{M}$ (sucrose), $C_2 = 0.008\,\text{M}$ (Na₂SO₄), and $i_1 = 1$ . The only unknown is $i_2$ , which depends on the degree of dissociation $\alpha$ .

Step‑by‑Step Solution

Write the isotonic condition Since $\Pi_{\text{sucrose}} = \Pi_{\text{Na₂SO₄}}$ and $R$ and $T$ cancel:

$i_{\text{sucrose}} \times 0.02 = i_{\text{Na₂SO₄}} \times 0.008$
With $i_{\text{sucrose}} = 1$ :
$0.02 = 0.008 \times i_{\text{Na₂SO₄}}$

Solve for the van’t Hoff factor of Na₂SO₄

$i_{\text{Na₂SO₄}} = \frac{0.02}{0.008} = 2.5$

Relate $i$ to degree of dissociation $\alpha$ For an electrolyte that dissociates into $n$ ions:

$i = 1 + (n - 1)\alpha$
Here, $\text{Na}_2\text{SO}_4$ gives $n = 3$ ions (2 Na⁺ + 1 SO₄²⁻). So:
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KCET 2025Set D-41 markMCQ
$Q. Among the following 0.1 m aqueous solutions, which one will exhibit the lowest boiling point elevation, assuming complete ionization of the compound in solution? (A) Aluminium chloride (B) Aluminmium sulphate (C) Potassium sulphate (D) Sodium chloride$
›Reveal solutionSolution
$\Delta T_b = i K_b m$ ; all solutions have the same $m$ and solvent, so the lowest elevation goes to the solute with the smallest van't Hoff factor $i$ — NaCl, with $i = 2$ .

Step 1 — The governing relation.

For elevation of boiling point of an electrolyte solution,
$\Delta T_b = i\,K_b\,m$
where $i$ = van't Hoff factor (number of particles produced per formula unit on complete ionisation), $K_b$ = molal elevation constant of the solvent (water, same for all), $m$ = molality ( $0.1\ \text{m}$ , same for all).

So $\Delta T_b \propto i$ .

Step 2 — Compute $i$ for each solute (complete ionisation assumed).

(A) Aluminium chloride: $\ce{AlCl3 -> Al^{3+} + 3Cl^-}$ ⇒ $i = 1 + 3 = \mathbf{4}$

(B) Aluminium sulphate: $\ce{Al2(SO4)3 -> 2Al^{3+} + 3SO4^{2-}}$ ⇒ $i = 2 + 3 = \mathbf{5}$

(C) Potassium sulphate: $\ce{K2SO4 -> 2K^+ + SO4^{2-}}$ ⇒ $i = 2 + 1 = \mathbf{3}$ …
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KCET 2025Set D-41 markMCQ
$Q. The electronic conductance depends on (A) Nature of electrolyte added (B) The number of valence electrons per atom (C) Concentration of the electrolyte (D) Size of the ions$
›Reveal solutionSolution
Distinguish electronic (metallic) conduction from electrolytic conduction: electronic conductance is a property of the metal's own free electrons, so it scales with the number of valence electrons each atom contributes.

Step 1 — What is electronic conductance?

There are two distinct conduction mechanisms:

Electronic (metallic) conduction — charge is carried by delocalised free electrons moving through the lattice of a metal or a semiconductor. No matter is transported.

Electrolytic (ionic) conduction — charge is carried by ions migrating through a solution or a molten salt. Matter is transported.

The question names electronic conductance, so anything referring to an electrolyte or to ions is off-mechanism by definition.

Step 2 — The factors electronic conductance actually depends on.

The nature and structure of the metal (how the lattice scatters electrons).

The number of valence electrons per atom — each atom donates its valence electrons to the electron "sea", so more valence electrons per atom means a larger free-electron density $n$ , hence a larger conductivity, since

$\sigma = n e \mu$
where $n$ is the free-electron number density, $e$ the electronic charge and $\mu$ the electron mobility. …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{K}_2 \mathrm{HgI}_4$
›Reveal solutionSolution
The van’t Hoff factor accounts for the number of particles formed per formula unit. For $\mathrm{K}_2 \mathrm{HgI}_4$ , complete dissociation gives 3 ions, but at 95% ionisation the effective factor is $1 + (3-1) \times 0.95 = 2.90$ , so the answer is (D).

The van’t Hoff factor $i$ tells us how many particles a solute actually produces in solution compared to the number of formula units dissolved. For a salt that dissociates, $i$ is larger than 1; for complete dissociation, $i$ equals the number of ions per formula unit. Here, the salt is only 95% ionised, so we need to find the effective number of particles.

Why this approach works:

If a compound dissociates into $n$ ions, and only a fraction $\alpha$ (the degree of ionisation) actually dissociates, then for every 1 mole of formula units, we get:

$(1 - \alpha)$ moles of undissociated compound,

$\alpha$ moles of each ion, but there are $n$ ions per formula unit, so total moles of ions = $n\alpha$ .

Total moles of particles in solution = $(1 - \alpha) + n\alpha = 1 + (n-1)\alpha$ .

That’s exactly the van’t Hoff factor: $i = 1 + (n-1)\alpha$ .

Now apply it step by step.

Identify the number of ions ( $n$ ) from complete dissociation. $\mathrm{K}_2 \mathrm{HgI}_4$ dissociates as:

$\mathrm{K}_2 \mathrm{HgI}_4 \rightarrow 2\mathrm{K}^+ + \mathrm{HgI}_4^{2-}$
That’s $2 + 1 = 3$ ions total. So $n = 3$ .

Use the given degree of ionisation.

$\alpha = 95\% = 0.95$ .

Plug into the formula.

$i = 1 + (n-1)\alpha = 1 + (3-1) \times 0.95 = 1 + 2 \times 0.95 = 1 + 1.90 = 2.90$ …
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COMEDK 2025Set 2025-M1 markMCQ
$9.2 \times 10^{-3} \mathrm{~kg}$
›Reveal solutionSolution

Molality $=0.333\ \text{m}$ , van't Hoff factor $i = 1.3$ (30% dissociation), so $\Delta T_f \approx 0.81\ \text{K}$ and the freezing point is $\approx 272.2\ \text{K}$ — option (B).

Moles and molality of formic acid
$n = \frac{9.2\ \text{g}}{46\ \text{g mol}^{-1}} = 0.2\ \text{mol}, \qquad m = \frac{0.2\ \text{mol}}{0.6\ \text{kg}} = 0.333\ \text{mol kg}^{-1}.$
van't Hoff factor — $\text{HCOOH} \rightarrow \text{H}^{+} + \text{HCOO}^{-}$ , so with degree of dissociation $\alpha = 0.30$ : …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{KCl}(\mathrm{M}=74.5 \mathrm{~g} \mathrm{~mol}^{-1})$
›Reveal solutionSolution
Isotonic solutions have the same osmotic pressure. For KCl, which dissociates, the van’t Hoff factor $i$ is found by equating its effective particle concentration to that of urea (a non‑electrolyte). The degree of dissociation $\alpha$ is then $i-1$ (since KCl gives 2 ions). The result is $\alpha = 0.96$ , option (A).

Concept & Intuition

Isotonic means the two solutions exert the same osmotic pressure at the same temperature. Osmotic pressure depends on the total number of particles in solution, not just the formula units. Urea does not dissociate, so its particle concentration equals its molar concentration. KCl, however, dissociates into K⁺ and Cl⁻, so each mole of KCl gives more than one mole of particles. The van’t Hoff factor $i$ accounts for this: $i = 1 + \alpha$ for a salt that gives two ions. By setting the osmotic pressures equal, we solve for $i$ and then for $\alpha$ .

Step‑by‑step solution

Write the osmotic pressure equality For two solutions at the same temperature, isotonic means $\pi_1 = \pi_2$ . Osmotic pressure $\pi = iCRT$ , where $C$ is molar concentration. Since $R$ and $T$ are identical, we have:

$i_{\text{KCl}} \, C_{\text{KCl}} = i_{\text{urea}} \, C_{\text{urea}}$
Urea is a non‑electrolyte, so $i_{\text{urea}} = 1$ .

Calculate molar concentrations For KCl: mass = $1.9\ \text{g}$ per $100\ \text{mL}$ = $0.1\ \text{L}$ . Molar mass $M_{\text{KCl}} = 74.5\ \text{g mol}^{-1}$ .

$C_{\text{KCl}} = \frac{1.9}{74.5 \times 0.1} = \frac{1.9}{7.45} \approx 0.2550\ \text{mol L}^{-1}$
For urea: mass = $3.0\ \text{g}$ per $100\ \text{mL}$ = $0.1\ \text{L}$ .

Molar mass $M_{\text{urea}} = 60.0\ \text{g mol}^{-1}$ .
$C_{\text{urea}} = \frac{3.0}{60.0 \times 0.1} = \frac{3.0}{6.0} = 0.5000\ \text{mol L}^{-1}$

Set up the isotonic condition

$i_{\text{KCl}} \times 0.2550 = 1 \times 0.5000$
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{CF}_3 \mathrm{COOH}$
›Reveal solutionSolution
The freezing point of a 1 molal aqueous solution is highest when the solute causes the smallest freezing-point depression. Since freezing-point depression depends on the number of particles (van’t Hoff factor $i$ ), the compound that dissociates least (weakest acid) gives the highest freezing point. Acetic acid ( $\mathrm{CH_3COOH}$ ) is the weakest acid here, so its solution has the highest freezing point — option (D).

Concept & Intuition

Freezing-point depression is a colligative property: it depends only on the number of solute particles in solution, not on their identity. For a given molality, the depression $\Delta T_f = i \cdot K_f \cdot m$ . Here $m = 1$ molal, so the solution with the smallest van’t Hoff factor $i$ (fewest particles) will have the smallest $\Delta T_f$ and thus the highest freezing point. All four compounds are carboxylic acids that can dissociate: $\mathrm{RCOOH} \rightleftharpoons \mathrm{RCOO^-} + \mathrm{H^+}$ . The stronger the acid, the more it dissociates, giving a larger $i$ (closer to 2). The weakest acid dissociates least, so $i$ is closest to 1, yielding the highest freezing point.

Step-by-step reasoning

Identify the trend in acid strength

The acidity of carboxylic acids is enhanced by electron-withdrawing groups (like halogens) on the carbon adjacent to the carboxyl group. The more electronegative and the greater the number of such groups, the stronger the acid.

$\mathrm{CF_3COOH}$ : three highly electronegative fluorines → very strong acid.

$\mathrm{CCl_3COOH}$ : three chlorines → strong, but less than the fluoro analogue.

$\mathrm{Cl_2CHCOOH}$ : two chlorines → moderately strong.

$\mathrm{CH_3COOH}$ : no electron-withdrawing groups → weakest acid.

Relate acid strength to van’t Hoff factor $i$

For a weak acid, $i = 1 + \alpha$ , where $\alpha$ is the degree of dissociation. A stronger acid has a larger $\alpha$ at the same concentration, so $i$ is larger.

$\mathrm{CF_3COOH}$ : nearly fully dissociated → $i \approx 2$ .

$\mathrm{CCl_3COOH}$ : high dissociation → $i$ close to 2.

$\mathrm{Cl_2CHCOOH}$ : moderate dissociation → $i$ between 1 and 2. …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. In the following question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below. Assertion (A): van't Hoff factor for a solution of benzoic acid in benzene is less than one. Reason(R): Benzoic acid undergoes dissociation in benzene and its calculated molecular mass using colligative properties is lower than its actual molecular mass. (A) Assertion (A) and Reason (R) are correct (B) Assertion (A) and incorrect and Reason (R) is correct (C) Assertion (A) is correct but Reason (R) is incorrect (D) Both Assertion (A) and Reason (R) are incorrect$
›Reveal solutionSolution
The van’t Hoff factor for benzoic acid in benzene is less than one because the acid dimerises (associates) in a non‑polar solvent, not dissociates. The reason given is false, so the correct choice is (C).

Concept & Intuition

The van’t Hoff factor $i$ measures how many particles a solute produces in solution compared to the number of formula units dissolved.

If a solute dissociates (splits into ions), $i > 1$ .

If a solute associates (forms dimers, trimers, etc.), $i < 1$ .

Benzoic acid in water dissociates slightly (giving $i > 1$ ), but in a non‑polar solvent like benzene, the carboxylic acid groups form strong hydrogen‑bonded dimers. This halves the number of particles, so $i \approx 0.5$ .

The reason in the question claims “dissociation” – that is the classic pitfall.

Step‑by‑step reasoning

What does the van’t Hoff factor tell us? For a solute that associates into $n$ molecules per aggregate, the observed colligative property (e.g., freezing‑point depression) is smaller than expected, so

$i = \frac{\text{observed number of particles}}{\text{formula units dissolved}} < 1.$
For benzoic acid in benzene, dimerisation is well‑known:
$2\,\text{C}_6\text{H}_5\text{COOH} \rightleftharpoons (\text{C}_6\text{H}_5\text{COOH})_2,$
so the effective particle count drops → $i < 1$ . Hence Assertion (A) is correct.

Examine the Reason (R).

Reason (R) says: “Benzoic acid undergoes dissociation in benzene and its calculated molecular mass using colligative properties is lower than its actual molecular mass.”

“Dissociation” means breaking into smaller particles (ions or fragments). That would raise the particle count, giving $i > 1$ and a lower apparent molar mass.

But in benzene, benzoic acid associates (dimerises), which lowers the particle count, giving $i < 1$ and a higher apparent molar mass. Therefore the reason is factually wrong: it describes the opposite process.

Check the logical link between A and R. …
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COMEDK 2024Set 2024-M1 markMCQ
$\text { If a compound } \mathrm{X}_3 \mathrm{Y} \text { is } 60 \% \text { ionised in aqueous medium, what is its van't Hoff factor value? }$
›Reveal solutionSolution
The van’t Hoff factor accounts for the number of particles a compound produces in solution. For $\mathrm{X}_3\mathrm{Y}$ that is 60% ionised, the factor is $i = 1 + \alpha(n-1) = 1 + 0.6(4-1) = 2.8$ , so the correct option is (D).

Concept and intuition

The van’t Hoff factor $i$ tells us how many moles of particles (ions or molecules) are actually present in solution per mole of compound dissolved. For a compound that ionises, $i$ is larger than 1; for one that associates, it is less than 1. The key formula is:
$n$
Here, $\mathrm{X}_3\mathrm{Y}$ breaks into 4 ions: $3\,\mathrm{X}^+$ and $1\,\mathrm{Y}^{3-}$ (or similar charges — the exact charges don’t matter for counting particles). So $n = 4$ . The degree of ionisation $\alpha = 60\% = 0.6$ . Plugging in gives $i = 1 + 0.6(4-1) = 1 + 0.6 \times 3 = 1 + 1.8 = 2.8$ .

Watch out

A common mistake is to forget that $n$ is the total number of ions produced, not just the number of different ions. Here, $\mathrm{X}_3\mathrm{Y}$ gives 3 X-ions and 1 Y-ion, so $n = 4$ , not 2.

Step-by-step reasoning

Identify the number of particles ( $n$ ) The compound $\mathrm{X}_3\mathrm{Y}$ dissociates as:

$\mathrm{X}_3\mathrm{Y} \rightarrow 3\mathrm{X}^+ + \mathrm{Y}^{3-}$
So one formula unit yields $3 + 1 = 4$ ions. Hence $n = 4$ .

Determine the degree of ionisation ( $\alpha$ )

The problem states it is 60% ionised, so $\alpha = 0.6$ .

Apply the van’t Hoff factor formula …
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COMEDK 2023Set 2023-E1 markMCQ
$3.42 \mathrm{~g}$
›Reveal solutionSolution

(Physically sensible: CaCl2 would give i = 3 if fully dissociated; 2.47 indicates incomplete dissociation / ion pairing.)

Concept: osmotic pressure of an electrolyte solution, pi = i C R T.

Molarity:

n(CaCl2) = 3.42 / 111 = 0.0308 mol

V = 2500 mL = 2.5 L

C = 0.0308 / 2.5 = 0.01233 M

van't Hoff factor:

i = pi / (C R T)

= 0.75 / (0.01233 x 0.0821 x 300)

= 0.75 / 0.3037

= 2.47 …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{X}$
›Reveal solutionSolution

The electrolyte $\mathrm{X_2A}$ gives 3 particles per formula unit ( $i=3$ ), so its osmotic pressure is $3\times0.04\,RT=0.12\,RT$ . The non-electrolyte at $0.2$ M gives $P_2=0.2\,RT$ . Their ratio is $0.12/0.2=0.6$ , so $\pi=0.6\,P_2$ .

Osmotic pressure $\pi = iCRT$ .

For X ( $\mathrm{X_2A}$ , complete dissociation):
$\mathrm{X_2A \to 2X^+ + A^{2-}},\quad i = 3$ $\pi = 3\times 0.04\times RT = 0.12\,RT$
For Y (non-electrolyte, $i=1$ ): …
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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

Concept and Intuition

Step‑by‑Step Solution

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 _____________.

Unlock everything free for 14 days