Question

An aqueous solution contains \(3.00 \%\) phenylalanine \(\left(\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{NO}_{2}\right)\) by mass. (Phenylalanine is nonionic and nonvolatile.) Find the following: (a) the freezing point of the solution (b) the boiling point of the solution (c) the osmotic pressure of the solution at \(25^{\circ} \mathrm{C}\)

Short answer

(a) -0.348°C (b) 100.096°C (c) 4.58 atm

Step-by-step solution

Understanding Given Information

The mass percentage of phenylalanine in the solution is given as 3.00%. We assume 100 g of solution, which results in 3 g of phenylalanine. The molar mass of phenylalanine is 165.19 g/mol. Thus, number of moles of phenylalanine = \( \frac{3\, \text{g}}{165.19\, \text{g/mol}} \approx 0.0181\, \text{mol} \).

Calculate Molality

The mass of water in the solution is 97 g or \( 0.097\, \text{kg} \). Molality \(m\) is calculated as \( m = \frac{0.0181\, \text{mol}}{0.097\, \text{kg}} \approx 0.187 \,\text{mol/kg}\).

Calculate Freezing Point Depression

The freezing point depression constant \( K_f \) for water is \( 1.86\, \text{K kg/mol} \). The freezing point depression \( \Delta T_f = K_f \times m = 1.86 \, \text{K kg/mol} \times 0.187 \,\text{mol/kg} \approx 0.348 \text{K} \). The freezing point of water is 0°C, so the freezing point of the solution is \( 0 - 0.348 \approx -0.348^{\circ}\text{C} \).

Calculate Boiling Point Elevation

The boiling point elevation constant \( K_b \) for water is \( 0.512\, \text{K kg/mol} \). The boiling point elevation \( \Delta T_b = K_b \times m = 0.512 \times 0.187 \approx 0.096 \text{K} \). Normal boiling point of water is 100°C, so the boiling point of the solution is \( 100 + 0.096 \approx 100.096^{\circ}\text{C} \).

Calculate Osmotic Pressure

For ideal solutions, osmotic pressure \( \Pi \) is given by \( \Pi = iMRT \), where \( i = 1 \) (for nonionic solute), \( M = 0.187 \text{ mol/L} \), \( R = 0.0821 \text{ L atm/mol K} \), and \( T = 298 \text{ K} \). Assuming solution volume is same as solvent volume, \( \Pi = 1 \times 0.187 \times 0.0821 \times 298 \approx 4.58 \text{ atm} \).

Key concepts

Freezing Point Depression

When a solute is added to a solvent, the solution's freezing point becomes lower than the pure solvent's. This concept, known as freezing point depression, occurs because the solute disrupts the solidification of the solvent molecules. In simpler terms, more energy (cooling) is required to arrange the solvent molecules into a solid structure.

Let's explore this through the example provided. When water, which usually freezes at 0°C, contains phenylalanine, a nonionic and nonvolatile compound, the freezing point is depressed by a certain degree due to the presence of 3.00% phenylalanine by mass. The change in freezing point is calculated using the formula:

• \( \Delta T_f = K_f \times m \)

where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the freezing point depression constant for water (1.86 K kg/mol), and \( m \) is the molality of the solution. Using these values in the provided solution, we calculated a freezing point depression of approximately 0.348°C. Thus, the new freezing point of the solution is about -0.348°C.

Understanding this concept helps us grasp how solutes affect phase transitions, showing the practical applications in fields like food preservation and antifreezing products.

Boiling Point Elevation

Boiling point elevation is the phenomenon where a solution's boiling point is higher than that of the pure solvent. This occurs because the solute particles increase the solvent's vapor pressure threshold needed to start boiling. In easy terms, more heat is required to allow the solvent to transition into a gas.

In the given example, by adding phenylalanine to water, the boiling point of water (normally 100°C) increased due to the boiling point elevation effect. The elevation in the boiling point can be determined using the formula:

• \( \Delta T_b = K_b \times m \)

Here, \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the boiling point elevation constant for water (0.512 K kg/mol), and \( m \) is again the molality of the solution. The calculated boiling point elevation from the exercise, approximately 0.096 K, results in the solution boiling at 100.096°C.

This knowledge is quite relevant industrially, particularly in cooking and purification processes, and helps explain why adding salt to water increases its boiling point, among other examples.

Osmotic Pressure

Osmotic pressure is an important colligative property that refers to the pressure needed to prevent the flow of solvent molecules through a semipermeable membrane into a more concentrated solution. It is a crucial concept in biological systems and chemical processes alike.

For our example, the osmotic pressure of the phenylalanine solution is calculated using the formula:

• \( \Pi = iMRT \)

In this formula, \( \Pi \) represents the osmotic pressure, \( i \) is the van't Hoff factor (which is 1 for nonionic solutes like phenylalanine), \( M \) is the molarity, \( R \) is the ideal gas constant (0.0821 L atm/mol K), and \( T \) is the temperature in Kelvin. From the task, the calculated osmotic pressure was approximately 4.58 atm at 25°C (298 K).

This measure of osmotic pressure is extremely relevant in various fields. It helps in understanding water retention in cells, nutrient absorption, and even the workings of blood pressure regulation, illustrating just how critical osmotic pressure is in real-world applications.