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

Chhattisgarh CgbseMCQ· 1mImportance★★★★★

59% · 77/131 Questions

🔒 Locked · start free trial →

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

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

Concept understanding — 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 $n$ . …

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

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: …

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

←Previous questionQ10 — The unit of ebullioscopic constant is ________. (i) Kkgmol⁻¹ or K(mola…→Next questionQ12 — An unripe mango placed in a concentrated salt solution to prepare pick…

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

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 _____________.

🎓Unlock everything free for 14 days

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