Question

An aqueous solution containing ionic salt having molality equal to \(0.1892\) freezes at \(-0.704^{\circ} \mathrm{C}\). The van't Hoff factor of the ionic salt will be equal to \(\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{Km}^{-1}\right)\)

Short answer

The van't Hoff factor is approximately 2.

Step-by-step solution

Understand the problem

We have an ionic solution with a given molality and freezing point depression, and we need to determine the van't Hoff factor. We use the formula for freezing point depression: \[ \Delta T_f = i \cdot K_f \cdot m \]where \( \Delta T_f \) is the depression in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality.

Identify known values

From the problem statement, we know:- \( \Delta T_f = 0.704^{\circ} C \)- \( K_f = 1.86 \mathrm{Km}^{-1} \)- \( m = 0.1892 \)

Apply the formula

Substitute the known values into the freezing point depression formula:\[ 0.704 = i \cdot 1.86 \cdot 0.1892 \]

Solve for the van't Hoff factor

Rearrange the formula to solve for \( i \):\[ i = \frac{0.704}{1.86 \times 0.1892} \]

Calculate the van't Hoff factor

Perform the calculation:\[ i = \frac{0.704}{1.86 \times 0.1892} \approx 2 \]

Key concepts

van't Hoff Factor

The van't Hoff factor, represented by the symbol \(i\), is a dimensionless quantity that gives us an understanding of how many particles a compound dissociates into when dissolved in a solution. It is crucial in determining colligative properties like freezing point depression.In the given problem, the van't Hoff factor is calculated to understand how an ionic salt behaves when it is dissolved. For ionic compounds, \(i\) reflects the number of ions generated. For instance, if a salt like sodium chloride (NaCl) dissociates completely into sodium (Na⁺) and chloride (Cl⁻) ions, the van't Hoff factor would be 2.Understanding \(i\) helps us to predict and compare the behavior of different solutions.- **Applications**: Useful in both theoretical calculations and practical applications such as determining electrolyte strength in solutions.
- **Note**: The value of \(i\) depends on the nature of the solute and the solvent’s abilities. It also depends on conditions such as concentration and temperature.

Molality

Molality is a way to express the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality is based on the mass of the solvent, making it independent of temperature changes.In the original exercise, molality is given as \(0.1892\), which indicates how concentrated the ionic salt is in the water. Knowing the molality of a solution is essential when calculating freezing point depression, as it allows us to understand how much the presence of a solute will lower the freezing temperature of the solution. Why choose molality over molarity?

• **Temperature Independence**: Because molality is based on mass, it doesn't change with temperature, unlike volume-based molarity.
• **Precision**: Vital for precise colligative properties calculations, like determining changes in boiling or freezing points.

Cryoscopic Constant

The cryoscopic constant, designated by \( K_f \), is specific to each solvent and indicates how much the freezing point of a solvent decreases per molal concentration of a solute. It is a crucial factor in calculating the freezing point depression using the van't Hoff factor.In the problem provided, \( K_f = 1.86 \, \mathrm{Km}^{-1} \) for water. This constant helps us to quantify how much the freezing point of water changes when a solute is added.The cryoscopic constant is essential for:

• **Calculating Freezing Point Depression**: It determines how a specific solute concentration impacts the freezing point.
• **Comparing Solvents**: Different solvents have different \( K_f \) values, influencing how they react to the same solute.
• **Understanding Properties of Solutions**: The value helps in modeling how solutions will behave under various conditions, which is key for applications in fields like chemistry and engineering.

By integrating \( K_f \) into the formula, students can accurately determine changes in freezing points that result from adding solutes to solvents.