Problem 43 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following \(0.1 \mathrm{M}\) aqueous solutions will have the lowest freezing point? (a) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{NaCl}\) (c) urea (d) glucose

Step-by-Step Solution

Answer

(a) \( \mathrm{K}_{2} \mathrm{SO}_{4} \) has the lowest freezing point.

Step 1: Understand the Concept

To determine which solution has the lowest freezing point, we need to use the concept of freezing point depression. The freezing point of a solution lowers as the number of dissolved particles increases due to colligative properties. The formula for freezing point depression is given by \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van 't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution. The larger the value of \( i \), the greater the freezing point depression.

Step 2: Determine the Van 't Hoff Factor

Calculate the van 't Hoff factor \( i \) for each compound:- \( \text{K}_2\text{SO}_4 \) dissociates into 3 ions: 2 \( \text{K}^+ \) and 1 \( \text{SO}_4^{2-} \), so \( i = 3 \).- \( \text{NaCl} \) dissociates into 2 ions: 1 \( \text{Na}^+ \) and 1 \( \text{Cl}^- \), so \( i = 2 \).- Urea and glucose do not dissociate into ions, so for both \( i = 1 \).

Step 3: Compare the Freezing Point Depressions

Since the molality and \( K_f \) (a property of the solvent, assumed the same for each) are the same for all solutions, the solution with the highest \( i \) will have the greatest freezing point depression. Thus:- \( \text{K}_2\text{SO}_4 \) has \( i = 3 \)- \( \text{NaCl} \) has \( i = 2 \)- Urea has \( i = 1 \)- Glucose has \( i = 1 \)Therefore, \( \text{K}_2\text{SO}_4 \) will have the lowest freezing point.

Step 4: Conclusion

With \( i = 3 \), \( \text{K}_2\text{SO}_4 \) produces the highest number of ions, leading to the largest freezing point depression, resulting in the lowest freezing point among the given options.

Key Concepts

Van 't Hoff Factor

The van 't Hoff factor, represented as \( i \), is crucial in understanding how solutes affect the freezing point of a solution. It indicates the number of particles a compound dissociates into in a solution. The more particles there are, the greater the effect on colligative properties like freezing point depression.

Take, for example, \( \text{K}_2\text{SO}_4 \). When dissolved, it separates into three ions: two potassium ions \( (\text{K}^+) \) and one sulfate ion \( (\text{SO}_4^{2-}) \). Therefore, \( i = 3 \).

For \( \text{NaCl} \), it dissociates into two ions, \( \text{Na}^+ \) and \( \text{Cl}^- \), so \( i = 2 \). Urea and glucose, however, do not dissociate into ions in solution, giving each an \( i = 1 \).

The van 't Hoff factor helps us predict which solutions will have a greater impact on the freezing point. A high \( i \) value indicates more particles and, therefore, a greater depression of the freezing point.

Colligative Properties

Colligative properties are traits of a solution that depend on the number of solute particles present, not their identity. They include boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression. The concept is straightforward: the more particles you introduce into a solvent, the more these properties are affected.

Freezing point depression is a prime example. When solute particles are added to a solvent, they disrupt the regular formation of the solid structure, requiring a lower temperature to freeze. It is calculated using the formula:

• \( \Delta T_f = i \cdot K_f \cdot m \)
• Where \( \Delta T_f \) is the change in freezing point, \( i \) is the van 't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality.

The key takeaway is that these properties aren't concerned with the type of solute but its quantity—more particles mean a more significant impact.

Molality

Molality is a measure of concentration that differs from molarity. It describes the moles of solute per kilogram of solvent. This distinction is vital since molality doesn't change with temperature. It's represented by \( m \) in the equation for freezing point depression.

• Formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)

Why does this matter? Because in calculations involving temperature changes, volume can vary due to expansion or contraction, affecting molarity. But since mass remains constant with temperature, molality is a more reliable measure.

In freezing point depression, equal molality across a series of solutions means one can directly compare the effect of the number of particles—\( i \)—on the freezing point. This way, whether you're adding salt (like \( \text{NaCl} \)) or a sugar equivalent (like glucose or urea), you can predict which makes the solution colder based on the particle count alone.