Question

Four solutions of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) with the concentrations \(0.1 \mathrm{~m}, 0.01 \mathrm{~m}, 0.001 \mathrm{~m}\) and \(0.0001 \mathrm{~m}\) are available. The maximum value of colligative property corresponds to : (a) \(0.0001 \mathrm{~m}\) solution (b) \(0.001 \mathrm{~m}\) solution (c) \(0.01 \mathrm{~m}\) solution (d) \(0.1 \mathrm{~m}\) solution

Short answer

The maximum value of colligative property corresponds to the 0.1 m solution.

Step-by-step solution

Understanding Colligative Properties

Colligative properties depend on the number of particles of solute present in the solution and not on the nature of the solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Since \(\mathrm{K}_2\mathrm{SO}_4\) dissociates into three ions in solution (2 K+ and 1 SO4^2-), the number of particles present in the solution increases, which enhances the colligative properties.

Calculating the Number of Particles in Each Solution

To find the solution with the maximum value of colligative property, calculate the total number of moles of particles produced by each solution upon dissociation. For \(\mathrm{K}_2\mathrm{SO}_4\), the dissociation is \(\mathrm{K}_2\mathrm{SO}_4 \rightarrow 2\mathrm{K}^+ + \mathrm{SO_4}^{2-}\). Thus, for every 1 mole of \(\mathrm{K}_2\mathrm{SO}_4\), 3 moles of particles are produced. Multiply the concentration of each solution by 3 to obtain the effective concentration in terms of particle number.

Comparing Effective Concentrations

Now, compare the effective concentrations of particles in all given solutions to determine the one with the maximum colligative property. The solution concentrations are: 0.1 m, 0.01 m, 0.001 m, and 0.0001 m. Multiplying each by 3 (since each unit of \(\mathrm{K}_2\mathrm{SO}_4\) produces 3 particles), we get: 0.3 m, 0.03 m, 0.003 m, and 0.0003 m, respectively.

Identifying the Maximum Colligative Property

By comparing the effective concentrations calculated in the previous step, the solution with the concentration of 0.3 m (which comes from the original 0.1 m solution) has the highest number of dissolved particles and thus exhibits the maximum colligative property.

Key concepts

Boiling Point Elevation

Boiling point elevation occurs when a solute is dissolved in a solvent, raising the temperature at which the liquid boils. Essentially, the added particles from the solute interfere with the ability of the solvent molecules to escape into the gas phase, which requires more energy (higher temperature) for the liquid to reach its boiling point. The extent of the boiling point elevation is directly proportional to the molal concentration of the solute particles in the solution.

The formula for boiling point elevation is given by \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor (number of particles the solute dissociates into), \( K_b \) is the ebullioscopic constant (a property of the solvent), and \( m \) is the molal concentration of the solution. In our example with \( \mathrm{K}_2\mathrm{SO}_4 \) solutions, the higher the concentration of the solution, the greater the boiling point elevation, leading to the 0.1 m solution having the most significant effect.

Freezing Point Depression

Freezing point depression is the lowering of a solvent's freezing point due to the addition of a solute. Similarly to boiling point elevation, the solute particles disrupt the formation of the solid structure of the solvent, requiring a lower temperature to freeze. The degree of freezing point depression is also related to the concentration of solute particles.

The relationship is given by \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant (a property of the solvent), and \( m \) is the molal concentration. With \( \mathrm{K}_2\mathrm{SO}_4 \), which has a van't Hoff factor of 3, the solution of 0.1 m will cause a larger decrease in the freezing point than the other solutions.

Osmotic Pressure

Osmotic pressure is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane, which naturally allows solvent but not solute particles to pass through. This colligative property is crucial in biological systems and industrial processes where the separation of solutes and solvents is needed.

Osmotic pressure (\( \Pi \) can be calculated using the formula \( \Pi = i \cdot M \cdot R \cdot T \), where \( M \) is the molarity of the solution, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( i \) is the van't Hoff factor. In the case of \( \mathrm{K}_2\mathrm{SO}_4 \), which dissociates into 3 ions, the osmotic pressure will be directly proportional to the concentration, making the 0.1 m solution have the highest osmotic pressure.

Vapor Pressure Lowering

Vapor pressure lowering is the reduction in a solvent’s vapor pressure due to the presence of a non-volatile solute. It happens because the solute particles occupy space at the liquid’s surface, making fewer solvent molecules escape to the vapor phase. To understand this, envision a crowded room with limited exits; it's harder for people to leave quickly when there's more in the way.

This can be expressed through Raoult's law, stating that the vapor pressure of the solvent over the solution is directly proportional to the mole fraction of the solvent. The formula is \( P = X_{solvent} \cdot P^0_{solvent} \), where \( P \) is the vapor pressure of the solvent in the solution, \( X_{solvent} \) is the mole fraction of the solvent, and \( P^0_{solvent} \) is the vapor pressure of the pure solvent. More solute particles mean lower vapor pressure. Following this, among the given \( \mathrm{K}_2\mathrm{SO}_4 \) solutions, the 0.1 m solution will exhibit the most substantial vapor pressure lowering.