Question

An aqueous solution of a solute which nerther assoc1ates nor dissociates has a freezing point depression of \(\mathrm{X}^{\circ} \mathrm{C}\). An equimolar solution of a second has a freezing point depression of \(4 \mathrm{X}^{\circ} \mathrm{C}\). The second solution could be a salt of formula (assuming \(100 \%\) dissociation ) (a) \(\mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{2}\) (c) \(\mathrm{A}_{3} \mathrm{~B}\) (d) \(\mathrm{AB}_{3}\)

Short answer

The salt could be any of \( \mathrm{A}_2 \mathrm{B}_2 \), \( \mathrm{A}_3 \mathrm{B} \), or \( \mathrm{AB}_3 \) since all dissociate into four ions.

Step-by-step solution

Understand the problem

We are given two solutions: one is a solute that neither associates nor dissociates, leading to a freezing point depression of \(X^\circ C\). The second solution is equimolar and causes a freezing point depression of \(4X^\circ C\), suggesting it dissociates into multiple ions. We need to determine the formula of the second salt based on its dissociation and resulting freezing point depression.

Determine the relationship between depression and number of ions

The freezing point depression \( \Delta T_f \) is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (number of particles the compound dissociates into), \( K_f \) is the freezing point depression constant, and \( m \) is the molality. An increase by a factor of 4 in freezing point depression suggests an increase in the number of ions \(4X = 4 \cdot K_f \cdot m \implies i = 4 \).

Identify the correct salt formula

We need the salt formula that dissociates into 4 ions. Checking each option:- For \(\mathrm{AB}_2\), it will dissociate into three ions: \(A^{+}\) and \(2B^{-}\), so \(i = 3\).- For \(\mathrm{A}_2\mathrm{B}_2\), it will dissociate into four ions: \(2A^{+}\) and \(2B^{-}\), so \(i = 4\).- For \(\mathrm{A}_3\mathrm{B}\), it will dissociate into four ions: \(3A^{+}\) and \(1B^{-}\), so \(i = 4\).- For \(\mathrm{AB}_3\), it will dissociate into four ions: \(A^{+}\) and \(3B^{-}\), so \(i = 4\).Options (b), (c), and (d) all dissociate into four ions.

Choose the correct option based on information given

Given that options (b), (c), and (d) all have the correct dissociation factor \(i = 4\), any of them could be correct as per the context provided by the freezing point depression being \(4X^\circ C\). Therefore, without additional information specifying which structure is expected, all three could meet the condition.

Key concepts

Van't Hoff Factor

The Van't Hoff factor, often represented by the symbol \(i\), is a way to express the effect of solute particles on colligative properties, such as freezing point depression. It represents the number of particles a solute forms when it dissolves in a solvent. For instance, if a salt dissociates completely, the factor reflects the total number of individual ions produced.

The reason this factor is important is due to its effect on physical properties of the solution, particularly those that depend solely on the number of particles in a solution, regardless of the type. In freezing point depression, the depression is directly proportional to the Van't Hoff factor, as shown in the equation, \[ \Delta T_f = i \cdot K_f \cdot m. \]

• \(\Delta T_f\) is the change in freezing point.
• \(K_f\) is the depression constant of the solvent.
• \(m\) is the molality of the solution.

For example, when a solute neither associates nor dissociates, \(i\) equals 1. However, if it dissociates into multiple particles, \(i\) is greater than 1, indicating an increased effect on freezing point depression.

Dissociation of Salts

Dissociation of salts involves the breaking apart of a compound into its individual ions when it dissolves in a solvent, such as water. This process is crucial to understanding colligative properties because the number of ions released impacts properties like freezing point depression.

For example, a salt like \(\text{NaCl}\) dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\), resulting in two ions and a Van't Hoff factor \(i = 2\). But, more complex salts, such as \(\text{A}_2\text{B}_2\), can yield even more ions, changing the factor to \(i = 4\) if it dissociates into four ions: two \(\text{A}^+\) and two \(\text{B}^-\).

The complete dissociation assumption often simplifies calculations and helps illustrate how different salts impact colligative properties. Without considering dissociation, predictions about freezing point changes would be inaccurate.

Colligative Properties

Colligative properties are those that depend on the number of particles in a solution, not the type of particles. This includes boiling point elevation, vapor pressure lowering, osmotic pressure, and most relevantly here, freezing point depression.

Freezing point depression occurs when a solute is added to a solvent, causing the freezing point to lower. This happens because the solute particles disrupt the formation of the structure necessary for the solvent to solidify, meaning more energy (lower temperatures) is needed to freeze.

• This effect is significantly impacted by the number of solute particles.
• For example, a solution with a solute that dissociates into five particles will depress the freezing point more than a solute dissociating into just two.

The equation \( \Delta T_f = i \cdot K_f \cdot m \) demonstrates this dependency, with \(i\) reflecting the role of dissociation. Thus, understanding colligative properties helps in predicting and interpreting experimental data concerning solutions.