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.

Showing the 12 most recent of 14 on this concept.

AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\mathrm{Na_2SO_4}$
›Reveal solutionSolution
Tests isotonicity via the van't Hoff factor for a dissociating electrolyte; the degree of dissociation of Na2SO4 works out to 75%.

Concept and Intuition

Two solutions are isotonic when they exert the same osmotic pressure — i.e. the same effective particle concentration. For an electrolyte that partially dissociates, the effective concentration is boosted by the van't Hoff factor $i = 1+(n-1)\alpha$ , where $n$ is the number of ions produced per formula unit and $\alpha$ is the degree of dissociation. A non-electrolyte like glucose has $i=1$ always.

Step-by-Step Solution

Glucose is a non-electrolyte: effective concentration $= 1\times 0.01 = 0.01$ M.

Isotonic condition: $i_{Na_2SO_4}\times C_{Na_2SO_4} = C_{glucose}$ , so $i\times 0.004 = 0.01 \Rightarrow i = 2.5$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$K_2SO_4$
›Reveal solutionSolution
Van't Hoff factor 2.70 for $K_2SO_4$ ( $n=3$ ions) gives $\alpha=0.85$ , so 15% remains undissociated.

Concept and Intuition

The van't Hoff factor $i$ measures how many effective particles a formula unit produces upon dissociation, relative to 1 (if it stayed intact). For a salt that dissociates into $n$ ions with degree of dissociation $\alpha$ , $i = 1 + (n-1)\alpha$ ; solving for $\alpha$ tells you what fraction actually ionised, and $1-\alpha$ is the undissociated fraction.

Step-by-Step Solution

$K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$ , so $n=3$ ions per formula unit.

$i = 1+(n-1)\alpha \Rightarrow 2.70 = 1+(3-1)\alpha = 1+2\alpha$ . …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$mol^{-1}$
›Reveal solutionSolution
Equal anti-freeze effect requires equal effective (van't Hoff-corrected) molality; NaCl's dissociation must be accounted for via $i$ , then glycerol (non-electrolyte) is matched to that effective concentration.

Concept and Intuition

Freezing-point depression is a colligative property: $\Delta T_f = i K_f m$ . For two solutions in the same solvent to show the same depression, their effective particle concentrations ( $i \times$ molality) must be equal — not their nominal concentrations. An electrolyte like NaCl contributes more particles per mole than a non-electrolyte like glycerol because it dissociates.

Step-by-Step Solution

NaCl is 97% dissociated into $Na^+ + Cl^-$ ( $\nu = 2$ ), so the van't Hoff factor is

$i = 1+\alpha(\nu-1) = 1+0.97(2-1) = 1.97$

Moles of NaCl in 1.0 L water: $n_{NaCl} = \dfrac{20}{58.5} = 0.3419$ mol (molar mass Na=23, Cl=35.5).

Effective moles (particles) contributed by NaCl: $i \times n = 1.97 \times 0.3419 = 0.6735$ mol. …
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AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$27^\circ C$
›Reveal solutionSolution
Rearranging the van't Hoff osmotic-pressure equation for $i$ and plugging in the numbers gives $i \approx 2.610$ .

Concept and Intuition

For an electrolyte solution, the observed osmotic pressure exceeds the "ideal" (non-dissociating) prediction because the solute dissociates into more particles. This is captured by the van't Hoff factor $i$ :
$\pi = i\,c\,R\,T$
where $c$ is the molar concentration, $R$ the gas constant, and $T$ the absolute temperature. $i>1$ signals dissociation (more particles than formula units), $i<1$ signals association.

Step-by-Step Solution

Given: $\pi = 0.65$ bar, $c = 0.01$ mol/L, $R = 0.083$ bar·L·K⁻¹·mol⁻¹, $T = 27+273 = 300$ K.

Compute the "ideal" osmotic pressure term: $cRT = (0.01)(0.083)(300) = 0.249$ bar. …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{CH_2FCOOH}$
›Reveal solutionSolution
Using the freezing-point depression formula with the van't Hoff factor gives $i\approx1.07$ , indicating the acid is a weak electrolyte that partially ionises.

Concept and Intuition

Colligative properties like freezing-point depression depend on the total number of solute particles in solution. For a solute that partially dissociates (like a weak acid), the van't Hoff factor $i$ (ratio of actual particles to formula units) is slightly greater than 1, capturing that partial ionisation.

Step-by-Step Solution

Molality: $m = \dfrac{n_{solute}}{\text{mass of solvent (kg)}} = \dfrac{0.25\,\mathrm{mol}}{0.5\,\mathrm{kg}} = 0.5\,\mathrm{mol/kg}$ .

Freezing point depression formula: $\Delta T_f = i\,K_f\,m$ . …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{CH_2FCOOH}$
›Reveal solutionSolution
The van't Hoff factor scales the ideal (non-dissociating) freezing-point depression up to the experimentally observed value: $\Delta T_f(\text{obs}) = i\,K_f\,m$ , giving $\approx 1.0$ K.

Concept and Intuition

For a weak electrolyte like fluoroacetic acid ( $\mathrm{CH_2FCOOH}$ ), partial dissociation produces slightly more particles than the un-ionised molality alone would suggest. The van't Hoff factor $i>1$ captures this extra particle count, and the observed colligative property is the ideal formula multiplied by $i$ .

Step-by-Step Solution

Formula: $\Delta T_f(\text{observed}) = i \times K_f \times m$ .

Plug in values: $i = 1.075$ , $K_f = 1.86$ K kg mol $^{-1}$ , $m = 0.5$ mol kg $^{-1}$ .

First compute the ideal part: $K_f \times m = 1.86 \times 0.5 = 0.93$ K. …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$\Delta T_b$
›Reveal solutionSolution
This tests the van't Hoff factor calculation from an observed elevation in boiling point of an electrolyte solution. The answer is (A) 1.92.

Concept and Intuition

The van't Hoff factor $i$ accounts for the actual number of particles a solute produces in solution relative to the ideal (undissociated) case. For a strong electrolyte like KCl, which fully dissociates into 2 ions, the ideal $i$ would be 2, but incomplete dissociation or ion-pairing effects can make the observed value slightly less than the ideal -- hence $i$ close to but below 2 for a dilute solution is physically reasonable.

Step-by-Step Solution

Colligative property formula for boiling point elevation with a van't Hoff factor: $\Delta T_b = i \cdot K_b \cdot m$ .

Rearranging: $i = \dfrac{\Delta T_b}{K_b \cdot m}$ . …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$\Delta T_f$
›Reveal solutionSolution
The van't Hoff factor is simply the ratio of the observed to the theoretical (undissociated, i=1) colligative property; here $i = 0.02/0.018 = 10/9$ .

Concept and Intuition

The van't Hoff factor $i$ measures how far a real solute's colligative behaviour deviates from the "ideal" (non-electrolyte, non-associating) prediction. $i>1$ signals dissociation (more particles than expected — e.g. acetic acid partially ionising into $H^+$ and $CH_3COO^-$ ), while $i<1$ signals association (fewer effective particles, e.g. dimerisation in benzene).

Step-by-Step Solution

"Calculated" $\Delta T_f$ = the theoretical value assuming the solute behaves as intact, non-dissociated molecules ( $i=1$ ): given as 0.018 K.

"Observed" $\Delta T_f$ = the actually measured value in the experiment: given as 0.02 K.

By definition, $i = \dfrac{\Delta T_f(\text{observed})}{\Delta T_f(\text{calculated}, i=1)} = \dfrac{0.02}{0.018}$ .

Simplify: $\dfrac{0.02}{0.018} = \dfrac{20}{18} = \dfrac{10}{9} \approx 1.11$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$Q. What is the van't Hoff factor of Ferric Sulphate (Assume 100% ionization) (A) 2 (B) 4 (C) 5 (D) 3$
›Reveal solutionSolution
Ferric sulphate $Fe_2(SO_4)_3$ dissociates completely into $2Fe^{3+} + 3SO_4^{2-}$ , five ions total, so with 100% ionization the van't Hoff factor $i = 5$ .

Concept and Intuition

The van't Hoff factor $i$ measures the ratio of the actual number of particles in solution to the number of formula units dissolved — it captures dissociation (for electrolytes, $i>1$ ) or association (for e.g. dimerizing acids in nonpolar solvents, $i<1$ ). For a strong electrolyte assumed to ionize completely, $i$ simply equals the total number of ions each formula unit breaks into.

Step-by-Step Solution

Write the complete dissociation of ferric sulphate: $Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$ .

Count the total ions produced per formula unit: $2\ (Fe^{3+}) + 3\ (SO_4^{2-}) = 5$ ions.

Since 100% ionization is assumed (given in the question), the van't Hoff factor equals this ion count exactly: $i = 5$ .

Common Mistakes …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$10^{-3}$
›Reveal solutionSolution
NaCl dissociates into 2 ions in water (i=2), so its freezing point depression is twice that of the non-dissociating glucose solution at the same molar concentration — giving −0.0372 °C.

Concept and Intuition

Freezing point depression depends on the total number of solute particles in solution, captured by the van't Hoff factor $i$ : $\Delta T_f = i\,K_f\,m$ . Glucose is a non-electrolyte ( $i=1$ ), while NaCl is a strong electrolyte that dissociates completely into Na⁺ and Cl⁻ ( $i=2$ ).

Step-by-Step Solution

From the glucose data, find $K_f$ : $\Delta T_f = K_f\times m \times i_{glucose}$ , with $i=1$ , $m=10^{-3}$ : $0.0186 = K_f\times10^{-3} \Rightarrow K_f = 18.6$ K·kg/mol. …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$BaCl_2$
›Reveal solutionSolution
Since all three solutions are 0.1 M, osmotic pressure ranks purely by the number of particles each solute produces in solution ( $i$ ), giving $BaCl_2 > NaCl >$ urea.

Concept and Intuition

Osmotic pressure is a colligative property: $\pi = iCRT$ . At the same molar concentration $C$ , temperature $T$ is fixed, so $\pi$ is directly proportional to the van't Hoff factor $i$ , which counts how many particles (ions) one formula unit produces on dissociation.

Step-by-Step Solution

Urea (B) is a non-electrolyte: it does not dissociate, so $i = 1$ .

NaCl (A) dissociates into $Na^+ + Cl^-$ : $i \approx 2$ .

$BaCl_2$ (C) dissociates into $Ba^{2+} + 2Cl^-$ (3 ions): $i \approx 3$ . …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$Q. The correct property of colloidal solutions when compared to that of true solutions is (A) Lower osmatic pressure (B) High depression in freezing point (C) High elevation in boiling point (D) High Vapour pressure$
›Reveal solutionSolution
This tests the comparative colligative behaviour of colloids vs true solutions — colloids show markedly lower osmotic pressure because particle number (not mass) governs colligative properties.

Concept and Intuition

Colligative properties (osmotic pressure, boiling point elevation, freezing point depression, vapour pressure lowering) depend on the NUMBER of solute particles, not their mass. Colloidal particles, though much bigger in size/mass, are far fewer in number per unit volume compared to a true (molecular) solution of the same mass concentration. Hence colloidal solutions show much smaller colligative effects — most notably a much lower osmotic pressure than expected, which is in fact used experimentally to estimate very large (macromolecular) molar masses.

Step-by-Step Solution

Colligative properties $\propto$ number of particles (moles) of solute per unit volume of solvent.

In a colloidal dispersion, particles (aggregates/micelles) are much bigger, so far fewer particles exist per gram of dispersed substance than in a true solution of the same mass.

Therefore, at the same mass concentration, the colloidal system exhibits a much LOWER osmotic pressure than the true solution. …
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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 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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