Q.The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is : ($K_f$ for water = 1·86 K kg $mol^{-1}$) (A) $-3·72\,°C$ (B) $+3·72\,°C$ (C) $-1·86\,°C$ (D) $+2·72\,°C$

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).

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:

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

2. 0.1 M Urea — Urea is also covalent and non-electrolytic. $i = 1$. Same depression as glucose.

3. 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. …

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:

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:

1. KCl (potassium chloride) This strong electrolyte dissociates completely:

Each formula unit produces 2 ions, so $i = 2$.

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

Each formula unit produces 3 ions, so $i = 3$.

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

Each molecule remains one particle, so $i = 1$.

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

Each formula unit produces 4 ions, so $i = 4$. …

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:

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.

1. 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”, $i = 2$.

2. Plug into the depression formula.

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 $\Delta T_f = 3.72\ ^\circ\text{C}$.

3. Apply the depression to the pure solvent’s freezing point. …

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.

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:

1. 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.

2. 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 $= (1 - \alpha) + \frac{\alpha}{n}$

3. 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:

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

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

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

4. Analyzing the Value of $i$ for Association: …

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:

Now, let's analyze the given assertion and reason.

1. 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.

2. 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.

3. Determine if Reason (R) is the correct explanation for Assertion (A). …