Question

Consider three test tubes. Tube A has pure water. Tube \(B\) has an aqueous \(1.0 \mathrm{~m}\) solution of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} .\) Tube \(\mathrm{C}\) has an aqueous \(1.0 \mathrm{~m}\) solution of \(\mathrm{NaCl}\). Which of the following statements are true? (Assume that for these solutions \(1.0 m=1.0 \mathrm{M} .)\) (a) The vapor pressure of the solvent over tube \(\mathrm{A}\) is greater than the solvent pressure over tube \(\mathrm{B}\). (b) The freezing point of the solution in tube \(\mathrm{B}\) is higher than the freezing point of the solution in tube \(\mathrm{A}\). (c) The freezing point of the solution in tube \(\mathrm{B}\) is higher than the freezing point of the solution in tube \(\mathrm{C}\). (d) The boiling point of the solution in tube \(\mathrm{B}\) is higher than the boiling point of the solution in tube \(\mathrm{C}\). (e) The osmotic pressure of the solution in tube \(\mathrm{B}\) is greater than the osmotic pressure of the solution in tube C.

Short answer

a) Vapor pressure of Tube A is higher than that of Tube B. b) Freezing point of Tube A is lower than that of Tube B. c) Freezing point of Tube B is higher than that of Tube C. d) Boiling point of Tube B is higher than that of Tube C. e) Osmotic pressure of Tube B is higher than that of Tube C. Answer: a) Vapor pressure of Tube A is higher than that of Tube B, and c) Freezing point of Tube B is higher than that of Tube C.

Step-by-step solution

(a) Comparing vapor pressures of Tube A and Tube B

Raoult's law states that the vapor pressure of a solvent in a solution is equal to the mole fraction of solvent times the vapor pressure of the pure solvent. In Tube A, the solvent is pure water, which means its mole fraction is 1. So, its vapor pressure will be the highest possible for water. In Tube B, there is a 1.0 M solution of ethanol (C2H5OH) in water. Ethanol is a non-ionizing solute, so the mole fraction of water will be less than 1. According to Raoult's law, the vapor pressure of the solvent in Tube B would be less than that of Tube A. So, the statement (a) is true.

(b) Comparing freezing points of Tube A and Tube B

Colligative properties state that the freezing point of a solution (a solvent with solute particles) is always lower than that of the pure solvent. In Tube A, we have pure water, and in Tube B, we have a solution of ethanol. Since there is a solute in Tube B, its freezing point will be lower than that of Tube A. So, the statement (b) is false.

(c) Comparing freezing points of Tube B and Tube C

Again, we will use colligative properties to analyze the freezing points. Tube B has a 1.0 M solution of ethanol, which is a non-ionizing solute, while Tube C has a 1.0 M solution of NaCl, which ionizes into Na+ and Cl- ions. So, the effective concentration of solute particles is 2.0 M in Tube C. According to colligative properties, the change in freezing point is proportional to the concentration of solute particles. Thus, the freezing point of Tube C will be lower than that of Tube B. So, the statement (c) is true.

(d) Comparing boiling points of Tube B and Tube C

The boiling point elevation is also a colligative property, and it states that the boiling point of a solution is higher than that of the pure solvent. In Tube B and Tube C, we have solutions of solutes in water. As mentioned before, the concentration of solute particles is effectively higher in Tube C due to the ionization of NaCl. The boiling point elevation is directly proportional to the concentration of solute particles, which means that the boiling point of Tube C will be higher than that of Tube B. So, the statement (d) is false.

(e) Comparing osmotic pressures of Tube B and Tube C

Osmotic pressure is another colligative property and is directly proportional to the concentration of solute particles. As discussed before, the concentration of solute particles in Tube C is effectively higher than that in Tube B due to the ionization of NaCl. Hence, the osmotic pressure of the solution in Tube C will be higher than that in Tube B. So, the statement (e) is false. In conclusion, statements (a) and (c) are true, while statements (b), (d), and (e) are false.

Key concepts

Raoult's Law

Raoult's law is foundational in understanding the behavior of solutions. It quantifies the relationship between the vapor pressure of a solvent in a solution and the presence of a solute. Simply put, Raoult's law states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent present in the mixture. To express it mathematically, if a solute is non-volatile and does not react chemically with the solvent, Raoult's law can be given by the formula \( P = X_{solvent} \cdot P^0_{solvent} \), 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.

Applying this principle to Tube A in our exercise, which contains pure water, the mole fraction of water is 1, implying the vapor pressure remains unchanged. Conversely, Tube B includes a solution of ethanol in water, decreasing the mole fraction of water and thus lowering the vapor pressure according to Raoult's law.

Freezing Point Depression

Freezing point depression is a colligative property that illustrates how the addition of solute to a solvent results in the lowering of the freezing point of that solution compared to the pure solvent. This phenomenon is directly proportional to the molar concentration of the solute particles in the solution. The key equation representing this concept is \( \Delta T_f = k_f \cdot m \), with \( \Delta T_f \) being the freezing point depression, \( k_f \) the cryoscopic constant (which is specific to each solvent), and \( m \) the molality of the solution.

In the context of our exercise, pure water in Tube A freezes at 0°C, but the ethanol solution in Tube B has a lower freezing point because ethanol disrupts the crystal structure formation of ice, hence lowering the temperature required for the water to solidify. This is why statement (b) in our problem set is false.

Boiling Point Elevation

Boiling point elevation is the process by which the boiling point of a solvent increases upon the addition of a solute. Like other colligative properties, it is a function of the solute concentration and doesn't depend on the chemical identity of the solute. The formula for calculating the change in boiling point is \( \Delta T_b = k_b \cdot m \), where \( \Delta T_b \) is the change in boiling point, \( k_b \) is the ebullioscopic constant and \( m \) is the molality of the solution.

In our exercise example, since Tube B and Tube C both contain solutions, their boiling points will be higher compared to pure water. However, due to the ionization of NaCl in Tube C, resulting in double the number of solute particles compared to the ethanol in Tube B, the boiling point elevation in Tube C will be greater.

Osmotic Pressure

Osmotic pressure can be understood as the pressure required to stop the flow of solvent molecules through a semipermeable membrane from a region of low solute concentration to a region of high solute concentration. It is directly related to the molarity of the solute in a solution, reflected by the formula \( \Pi = MRT \), where \( \Pi \) represents the osmotic pressure, \( M \) is the molarity of the solution, \( R \) is the universal gas constant, and \( T \) is the absolute temperature.

When applying this to our Tubes B and C, the osmotic pressure in Tube C, which has a NaCl solution, will be greater than the ethanol solution in Tube B. This occurs because NaCl dissociates into two particles (Na+ and Cl-), increasing the total number of solute particles, and thereby increasing the osmotic pressure.

Solute-Solvent Interactions

Solute-solvent interactions are at the heart of understanding solution chemistry. These interactions determine the solubility, physical properties, and behaviors of solutions. In the absence of chemical reactions between solute and solvent, the strength and type of intermolecular forces—such as hydrogen bonding, dipole-dipole interactions, and dispersion forces—play a crucial role in how substances dissolve.

Considering our exercise, ethanol in Tube B forms hydrogen bonds with water, which is a major factor in its solubility. In contrast, the ionic bonds between Na+ and Cl- in Tube C break in the presence of water, a polar solvent, resulting in the separation of ions and solvation. The interplay of these forces also affects the vapor pressure, freezing point, and other colligative properties observed in our exercise's example.