Problem 167 - A Complete Resource Book in Chemistry for JEE Main

Question

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water \(\left(\Delta \mathrm{T}_{\mathrm{p}}\right)\), when \(0.01 \mathrm{~mol}\) of sodium sulphate is dissolved in \(1 \mathrm{Kg}\) of water, is \(\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\) (a) \(0.0372 \mathrm{~K}\) (b) \(0.0558 \mathrm{~K}\) (c) \(0.0744 \mathrm{~K}\) (d) \(0.0186 \mathrm{~K}\)

Step-by-Step Solution

Answer

The change in freezing point is 0.0558 K, answer (b).

Step 1: Determine the Formula for Freezing Point Depression

The formula for the freezing point depression is given by \[ \Delta T_f = i \cdot m \cdot K_f \]where: - \( \Delta T_f \) is the change in freezing point, - \( i \) is the van't Hoff factor (number of particles the compound dissociates into),- \( m \) is the molality of the solution,- \( K_f \) is the cryoscopic constant.

Step 2: Calculate the Van't Hoff Factor (i)

Sodium sulphate \( \text{Na}_2\text{SO}_4 \) dissociates completely in water to form 2 \( \text{Na}^+ \) ions and 1 \( \text{SO}_4^{2-} \) ion, totaling to 3 ions. Thus, the van't Hoff factor \( i = 3 \).

Step 3: Determine the Molality (m) of the Solution

Molality \( m \) is calculated as moles of solute per kilogram of solvent. Here, we have \( 0.01 \) moles of sodium sulphate in \( 1 \) kg of water. Thus, \( m = 0.01 \ \text{mol/kg} \).

Step 4: Calculate the Freezing Point Depression

Substitute the known values into the freezing point depression formula:\[\Delta T_f = i \cdot m \cdot K_f = 3 \times 0.01 \ \text{mol/kg} \times 1.86 \ \text{K kg/mol} = 0.0558 \ \text{K}\]Thus, the change in freezing point is \( 0.0558 \ \text{K} \).

Step 5: Identify the Correct Answer

From the calculated \( \Delta T_f = 0.0558 \ \text{K} \), the correct answer is option (b).

Key Concepts

Freezing Point Depression

Freezing point depression is one of the fascinating colligative properties observed in solutions. This phenomenon occurs when a solute is added to a solvent, resulting in a lower freezing point than the pure solvent. In simpler terms, the solution remains in a liquid state at temperatures where the pure solvent would freeze. This is because the solute particles disrupt the formation of a solid crystal lattice, which is necessary for freezing.

To quantify this change, we use the formula for freezing point depression: \[ \Delta T_f = i \cdot m \cdot K_f \]

• \( \Delta T_f \) (Change in freezing point): Shows how much the freezing point changes.
• \( i \) (van't Hoff factor): Represents the number of particles the solute turns into.
• \( m \) (Molality): The concentration of the solute in the solution.
• \( K_f \) (Cryoscopic constant): Represents the freezing point depression strength of the solvent.

Understanding and applying freezing point depression can help predict the behavior of various solutions, such as antifreeze in car engines.

Van't Hoff Factor

The van't Hoff factor, denoted by \( i \), is crucial in determining colligative properties of a solution. It quantifies the effect of solute particles in the solution. For instance, when ionic compounds dissolve, they break down into multiple ions. These ions contribute to lowering the freezing point moniker, thus amplifying the colligative effect.

Consider sodium sulfate \( Na_2SO_4 \). In water, it dissociates into:

• 2 sodium ions \( \text{Na}^+ \)
• 1 sulfate ion \( \text{SO}_4^{2-} \)

In total, this results in three particles, so the van't Hoff factor \( i \) is 3. The greater the value of \( i \), the more significant the effect on the freezing point depression and other such properties.

Molality

Molality plays a pivotal role in understanding colligative properties. It's a way to define the concentration of a solution, focusing on the solvent's mass instead of its volume. Molality is calculated as moles of solute per kilogram of solvent \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \).

This concentration measure is beneficial, especially in temperature-dependent studies, because unlike volume, the mass doesn't fluctuate with temperature. In the given example, \( 0.01 \) moles of sodium sulfate dissolved in \( 1 \) kg of water gives a molality \( m = 0.01 \, \text{mol/kg} \). This straightforward calculation supports comprehensive understanding in the study of solutions.

Cryoscopic Constant

The cryoscopic constant, \( K_f \), is a specific property related to a solvent's ability to undergo freezing point depression. Each solvent has its unique \( K_f \) value, which is expressed in \( \text{K} \cdot \text{kg/mol} \). This constant provides a numerical measure of how the presence of a solute impacts the solvent's freezing point.

Consider water's cryoscopic constant, \( K_f = 1.86 \, \text{K kg/mol} \), as used in our example. This means that per mole of solute added to 1 kg of water, the freezing point will decrease by \( 1.86 \, \text{K} \), assuming no dissociation or association. When combined with other factors like molality and the van't Hoff factor, \( K_f \) allows for precise calculation of a solution's new freezing point. Understanding \( K_f \) allows chemists to manipulate and predict solution behaviors reliably.