Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

KBr is \(80 \%\) dissociated in aqueous solution of \(0.5 \mathrm{M}\) concentration. (Given \(\mathrm{K}_{t}\) for water \(=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) ). The solution freezes at (a) \(271.326 \mathrm{~K}\) (b) \(272 \mathrm{~K}\) (c) \(270.5 \mathrm{~K}\) (d) \(268.5 \mathrm{~K}\)

Step-by-Step Solution

Answer

The solution freezes at (a) 271.326 K.

Step 1: Understand the Dissociation

KBr dissociates into 1 K\(^+\) ion and 1 Br\(^-\) ion: \(\text{KBr} \rightarrow \text{K}^+ + \text{Br}^-\). It is 80% dissociated; thus, 80% of the initial KBr dissociates into ions.

Step 2: Calculate the Van't Hoff Factor

The Van’t Hoff factor (i) accounts for the number of particles formed. For complete dissociation, \(i = 2\). Since only 80% dissociates, we use \[i = 1 + \alpha (n-1)\] where \(\alpha = 0.8\) (dissociation degree) and \(n=2\). Thus, \[i = 1 + 0.8(2-1) = 1.8\].

Step 3: Determine the Freezing Point Depression Formula

Use the formula: \(\Delta T_f = i \cdot K_t \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor, \(K_t = 1.86\, \mathrm{K} \, \mathrm{kg} / \mathrm{mol}\), and \(m = 0.5\, \mathrm{mol} / \mathrm{kg}\) is the molal concentration.

Step 4: Calculate the Freezing Point Depression

Substitute the values into the equation: \[\Delta T_f = 1.8 \cdot 1.86 \, \mathrm{K} \, \mathrm{kg} \, \mathrm{mol}^{-1} \cdot 0.5 \, \mathrm{mol} \, \mathrm{kg}^{-1}\] which yields \(\Delta T_f = 1.674\, \mathrm{K}\).

Step 5: Find the Freezing Point of the Solution

Water freezes at 273.15 K. The new freezing point of the solution is \(273.15 \, \mathrm{K} - 1.674 \, \mathrm{K} = 271.476 \, \mathrm{K}\).

Step 6: Compare and Conclude

Compare the calculated freezing point of 271.476 K with the provided options: (a) 271.326 K, (b) 272 K, (c) 270.5 K, (d) 268.5 K. The closest option is (a) 271.326 K.

Key Concepts

Van't Hoff Factor

The Van't Hoff factor, symbolized as \(i\), is crucial in understanding colligative properties, like freezing point depression. It signifies the number of particles a solute splits into when dissolved.
For instance, if a compound dissociates fully, the Van’t Hoff factor equals the number of total ions formed. For potassium bromide (KBr), which dissociates into \( \text{K}^+ \) and \( \text{Br}^- \), the theoretical \(i\) is 2 if fully dissociated.
However, real-world scenarios often feature partial dissociation, which needs adjustment. This is where the dissociation degree (\( \alpha \)) comes in.

• Complete dissociation: \( i = n \) (where \( n \) is the number of ions)
• Partial dissociation: \( i = 1 + \alpha (n - 1) \)

For our KBr example, with 80% dissociation \( (\alpha = 0.8) \), we calculate \( i = 1 + 0.8 \times (2 - 1) \), resulting in \( i = 1.8 \).
Understanding this factor is pivotal, not just theoretically, but for accurately predicting how compounds behave when dissolved and how they affect solution properties.

Dissociation

Dissociation in chemistry refers to the process where compounds break apart into smaller entities, usually ions, when dissolved in solvents like water. This transformation is central to understanding solution behavior, especially for ionic compounds.
Consider potassium bromide (KBr). In water, it dissociates into potassium ions \((\text{K}^+)\) and bromide ions \((\text{Br}^-)\). The dissociation is represented by the equation:

• \(\text{KBr} \rightarrow \text{K}^+ + \text{Br}^-\)

The degree of dissociation, \( \alpha \), quantifies how much of this process occurs. A high \( \alpha \) signifies more dissociation, leading to more free ions.

• For instance, if \( \alpha = 0.8 \), 80% of the solute dissociates.

The extent of dissociation influences the Van't Hoff factor, which in turn affects colligative properties like freezing point depression.
Accurately gauging dissociation helps predict and explain the behavior of solutions in various conditions.

Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the number of solute particles in a solution, rather than their identity. When a solute like an ionic compound dissolves in a solvent, it lowers the freezing point of the liquid.
The core principle is that the dissolved particles disrupt the formation of a solid lattice, requiring a lower temperature to solidify. The extent to which the freezing point is lowered is calculated using:

• \(\Delta T_f = i \cdot K_t \cdot m\)

where:

• \(\Delta T_f\) is the freezing point depression,
• \(i\) is the Van't Hoff factor,
• \(K_t\) is the cryoscopic constant (specific to the solvent), and
• \(m\) is the molality of the solution.

In our example, KBr in a 0.5 M aqueous solution causes the freezing point to drop.

• Calculate \(\Delta T_f = 1.8 \times 1.86 \times 0.5 = 1.674 \text{ K}\).
• The solution freezes at \( 273.15 - 1.674 = 271.476 \text{ K} \).

Recognizing how solutes alter freezing points is not only fascinating but also practical in fields like chemistry and material science, with real-world applications in antifreeze formulations and food preservation.