Problem 146 - A Complete Resource Book in Chemistry for JEE Main

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 ____ .\(\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{Km}^{-1}\right)\)

Step-by-Step Solution

Answer

The van't Hoff factor is approximately 2.00.

Step 1: Identify the Given Values

We have the following values:- Molality of the solution: \( m = 0.1892 \, \text{mol/kg} \)- Freezing point of the solution: \( T_f = -0.704^{\circ} \text{C} \)- Freezing point constant: \( K_f = 1.86 \, \text{Km}^{-1} \)The normal freezing point of water is \( 0^{\circ} \text{C} \). The freezing point depression \( \Delta T_f \) can thus be calculated as \( \Delta T_f = 0^{\circ} \text{C} - (-0.704^{\circ} \text{C}) = 0.704^{\circ} \text{C} \).

Step 2: Understand the Freezing Point Depression Formula

The formula for freezing point depression is: \\[ \Delta T_f = i \cdot K_f \cdot m \] \Where:- \( \Delta T_f \) is the freezing point depression- \( i \) is the van't Hoff factor- \( K_f \) is the cryoscopic constant- \( m \) is the molality of the solution

Step 3: Substitute the Known Values into the Formula

Substitute the known values into the freezing point depression formula: \\[ 0.704 = i \cdot 1.86 \cdot 0.1892 \] \This equation will allow us to solve for \( i \), the van't Hoff factor.

Step 4: Solve for the Van't Hoff Factor

Rearrange the equation to find \( i \): \\[ i = \frac{0.704}{1.86 \cdot 0.1892} \] \Calculate the above expression: \\[ i \approx \frac{0.704}{0.351912} \approx 2.00 \]

Step 5: Interpret the Result

Thus, the van't Hoff factor \( i \) is approximately \( 2.00 \). This indicates that the ionic salt dissociates into two ions upon dissolving in water.

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating property of solutions. This concept refers to the process where the freezing point of a liquid (in this case, water) is lowered by adding a solute (such as an ionic salt).
When substances like salt are dissolved in water, the intermolecular forces holding the water molecules together are disrupted.
This disruption means that a lower temperature is required to solidify the water, thus leading to a depression of the freezing point. It's critical to remember that the freezing point of the pure solvent must be higher than that of the solution. Here’s how this works for our example:

• The solution freezes at -0.704°C, while the pure solvent, water, typically freezes at 0°C.
• The difference, 0.704°C, represents the freezing point depression, denoted as ΔT_f.

This property is particularly useful in various practical applications, such as road salting in winter or making ice creams.

Cryoscopic Constant

The cryoscopic constant, represented as K_f, is a crucial factor in understanding freezing point depression. This constant is specific to each solvent and represents how effective a solute can be in lowering the freezing point of the solvent.
Think of K_f as the sensitivity measure of a liquid’s freezing shift per molal unit of solute. In the context of our problem:

• The cryoscopic constant for water is given as 1.86 Km^{-1}.
• This means that for every molality unit of a solute added to water, the freezing point is expected to drop by about 1.86°C, but only in an ideal situation.

In practice, the pure calculations might differ slightly due to factors like the dissociation of ionic compounds. But understanding K_f is essential for predicting and calculating the depression of freezing points accurately.

Molality

Molality is an important way to measure concentration in chemistry, especially when dealing with temperature changes. Measuring the amount of substance per mass unit of the solvent, it provides a consistent way to express solution concentration.
Why is molality often preferred over molarity in such cases? Mainly because molality does not change with temperature, unlike molarity which depends on the volume of the solution.
In our exercise:

• The molality of the ionic salt solution was 0.1892 mol/kg.
• This means for every kilogram of water, 0.1892 moles of the ionic salt were dissolved.

This consistency makes molality a useful measure in colligative properties, helping to predict how a solution will behave under different conditions of temperature and pressure.