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Consider \(0.1 \mathrm{M}\) solutions of two solutes \(X\) and \(Y\). The solute \(X\) behaves as a univalent electrolyte while the solute \(Y\) dimerises in solution. Which of the following statements are correct regarding these solutions? (1) The boiling point of the solution of \(X\) will be higher than that of \(Y\) (2) The osmotic pressure of the solution of \(Y\) will be lower than that of \(X\) (3) The freezing point of the solution of \(X\) will be lower than that of \(Y\) (4) The relative lowering of vapour pressure of both the solutions will be the same Select the correct answer from the option given below : (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,2 and 4 (d) 1,3 and 4

Short Answer

Expert verified
Statements 1, 2, 3, and 4 are correct; therefore the correct answer is (d) 1,3 and 4.

Step by step solution

01

Understanding Colligative Properties

Colligative properties are properties of solutions that depend on the number of particles in a given volume of solvent and not on the type of chemical species present. The properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. A univalent electrolyte like solute X will dissociate into two particles, while the dimerising solute Y will form molecules consisting of two Y units, effectively reducing the number of particles.
02

Comparing Boiling Point Elevation

Boiling point elevation is directly proportional to the molal concentration of the solution particles. Solute X will dissociate into two ions increasing the particle concentration and thereby increasing the boiling point more than solute Y, which dimerises and reduces its particle concentration in solution.
03

Analyzing Osmotic Pressure

Osmotic pressure is also a colligative property and depends on the molarity of the solute particles. Since solute X dissociates into more particles compared to solute Y, the solution of X will have a higher osmotic pressure than that of Y.
04

Determining Freezing Point Depression

Freezing point depression, like boiling point elevation, is dependent on the number of particles in solution. Solute X again leads to a higher concentration of particles, hence it will have a greater freezing point depression compared to the dimerised solution Y.
05

Comparing Relative Lowering of Vapor Pressure

The relative lowering of vapor pressure is directly proportional to the mole fraction of the solute particles. Assuming partial dissociation or dimerisation, the mole fraction of the solute particles remains the same in both solutions. Thus, the relative lowering of vapour pressure is the same for both X and Y.
06

Selection of Correct Answer

Since statements 1, 2, and 3 are correct as X will have a higher boiling point, higher osmotic pressure and lower freezing point in comparison to Y, and 4 is correct as the relative lowering of vapor pressure will be the same for both solutions, the correct answer is that all given statements (1, 2, 3, and 4) are true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boiling Point Elevation
Boiling point elevation is a colligative property which explains how a solvent's boiling point is raised when a solute is dissolved in it. The principle behind this phenomenon is that the presence of solute particles in a solution lowers the solvent's vapor pressure, meaning that a higher temperature is required for the vapor pressure to reach atmospheric pressure, which is the condition for boiling. The greater the number of dissolved particles, the higher the boiling point elevation.

For instance, in a solution with a univalent electrolyte, like the solute X in the exercise, the electrolyte will dissociate into individual ions. Each ion counts as a separate particle in the solution, effectively increasing the particle count and consequently the boiling point elevation. As a result, solution X's boiling point is higher than that of solution Y, where the solute dimerises, decreasing the number of particles in solution.
Osmotic Pressure
Osmotic pressure is another colligative property that is greatly influenced by the number of solute particles in a solution. It is the pressure required to prevent the inward flow of solvent molecules through a semipermeable membrane into a solution. When there are more solute particles, like in the case with solute X which dissociates into ions, the osmotic pressure will be higher compared to a solution where the solute, such as Y, dimerises and reduces the number of particles.

Understanding osmotic pressure is crucial because it helps explain how water moves across cell membranes, a fundamental process in biology known as osmosis. An increase in the concentration of solute particles will draw more water across the membrane, raising the osmotic pressure.
Freezing Point Depression
Freezing point depression is the process where the freezing point of a solvent is lowered by adding a solute. Like boiling point elevation, this property also relies on the number of solute particles. The presence of solute particles disrupts the solidification process of the solvent, requiring a lower temperature to freeze the solution.

In our specific case, the presence of the solute X, which dissociates, creates more solute particles, leading to a greater freezing point depression than in the solution of Y where the solute forms dimers (which counts as fewer particles). Hence, the solution of X will have a lower freezing point than that of Y. Freezing point depression is widely applied in real-world scenarios, such as using salt to melt ice on roads during winter.
Vapor Pressure Lowering
Vapor pressure lowering is a colligative property describing the decrease in vapor pressure when a non-volatile solute is dissolved in a solvent. The addition of a solute reduces the number of solvent molecules at the surface of the liquid that can escape into the gas phase, leading to a lower vapor pressure. The effect on vapor pressure is directly proportional to the mole fraction of the solute particles within the solution.

In the context of solutions X and Y, even though X has a higher number of particles due to dissociation and Y has a lower number due to dimerisation, the vapor pressure lowering will be the same for both solutions if their mole fractions are equal. This property is crucial for understanding and designing systems such as refrigeration cycles and predicting the behavior of solutions under different conditions.

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Most popular questions from this chapter

If \(M_{\text {normal }}\) is the normal molecular mass and \(\alpha\) is the degree of ionization of \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), then the abnormal molecular mass of the complex in the solution will be: (a) \(M_{\text {normal }}(1+2 \alpha)^{-1}\) (b) \(M_{\text {normal }}(1+3 \alpha)^{-1}\) (c) \(M_{\text {normal }}(1+\alpha)^{-1}\) (d) equal to \(M_{\text {normal }}\)

Assuming complete ionisation, the solution having maximum freezing point will be: (a) \(1 \mathrm{M} \mathrm{CaF}_{2}\) (b) \(1.5 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (c) \(2 \mathrm{M} \mathrm{NaCl}\) (d) \(1 \mathrm{M} \mathrm{AgNO}_{3}\)

The osmotic pressures of equimolar solutions of urea, \(\mathrm{BaCl}_{2}\) and \(\mathrm{AlCl}_{3}\) will be in the order : (a) \(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\) urea (b) \(\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}>\) urea (c) urea \(>\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}\) (d) \(\mathrm{BaCl}_{2}>\) urea \(>\mathrm{AlCl}_{3}\)

\(\mathrm{C}_{6} \mathrm{H}_{6}\) freezes at \(5.5^{\circ} \mathrm{C}\). At what temperature will a solution of \(10.44 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{10}\) in \(200 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\) freeze? \(K_{f}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=5.12^{\circ} \mathrm{C} / \mathrm{m}\) (a) \(4.608^{\circ} \mathrm{C}\) (b) \(0.892^{\circ} \mathrm{C}\) (c) \(5.5^{\circ} \mathrm{C}\) (d) none of these

The osmotic pressures of \(0.010 \mathrm{M}\) solutions of \(\mathrm{KI}\) and of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) are \(0.432\) atm and \(0.24\) atm respectively. The van't Hoff factor for KI is : (a) \(1.80\) (b) \(0.80\) (c) \(1.2\) (d) \(1.0\)

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