Question

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing 1.0 \(\mathrm{g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{g} / \mathrm{mol} )\) if the density of the solution is 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

Short answer

The freezing-point depression for the given solution is \(2.1 \times 10^{-5}\) °C and the osmotic pressure is 0.0027 atm. Osmotic pressure would be a better colligative property to determine the molar masses of large molecules, as the freezing-point depression value is too small and prone to measurement errors, while osmotic pressure provides a more significant value and is generally more accurate for large molecules.

Step-by-step solution

Calculate the molality of the protein in the solution

First, we need to find the molality (moles of solute per kilogram of solvent) of the protein in the solution. To do this, we can use the following formula: Molality = moles of solute / mass of solvent (in kg) We are given the mass concentration of protein in the solution (1.0 g/L). We need to convert it into moles by using its molar mass (9.0 x 10^4 g/mol): moles of protein = mass of protein / molar mass of protein moles of protein = (1.0 g) / (9.0 x 10^4 g/mol) = \(1.11 \times 10^{-5}\) mol We also need to find the mass of the solvent (water) in the solution. Since the density of the solution is given as 1.0 g/cm^3, we can assume that the mass of 1 L (1000 cm^3) of the solution would be 1000 g. Since the mass of protein is only 1 g, the mass of water in 1 L of the solution is about 999 g or 0.999 kg. Now, we can calculate the molality of the protein: Molality = moles of protein / mass of water Molality = \(1.11 \times 10^{-5}\) mol / 0.999 kg = \(1.11 \times 10^{-5}\) mol/kg

Calculate the freezing-point depression

Freezing-point depression is given by the formula: ΔT_f = K_f × molality where K_f is the freezing-point depression constant for water (1.86 °C/molal) Now, we can substitute the values and find the freezing-point depression: ΔT_f = (1.86 °C/molal) × \(1.11 \times 10^{-5}\) mol/kg ΔT_f = \(2.1 \times 10^{-5}\) °C

Calculate the osmotic pressure

Osmotic pressure is given by the formula: π = n × R × T / V where n is the number of moles of solute, R is the ideal gas constant (0.0821 L atm/(mol K)), T is the temperature in kelvins (25°C = 298 K), and V is the volume of the solution in liters (1 L). Now, substituting the values, we can find the osmotic pressure: π = \(1.11 \times 10^{-5}\) mol × (0.0821 L atm/(mol K)) × 298 K / 1.0 L π = 0.0027 atm Now, let us answer part (b) of the question.

Determine the better colligative property

Based on the values found for freezing-point depression (ΔT_f = \(2.1 \times 10^{-5}\) °C) and osmotic pressure (π = 0.0027 atm), it can be concluded that osmotic pressure would be a better colligative property to determine the molar masses of large molecules. The reason is that the ΔT_f is too small, which can lead to significant errors when measuring the freezing-point depression. On the other hand, osmotic pressure gives us a more significant value, and its measurement is usually more accurate for large molecules. Moreover, osmotic pressure measurement does not involve a phase change, which could lead to problems while measuring the freezing-point depression.

Key concepts

Freezing-Point Depression

Freezing-point depression is a colligative property that refers to the lowering of a solvent's freezing point due to the addition of a solute. This phenomenon occurs because the solute particles disrupt the formation of the solid structure of the solvent. The degree of freezing-point depression is directly related to the molality of the solution.

• The formula to calculate freezing-point depression is given by: \[ \Delta T_f = K_f \times \text{molality} \]
• \( K_f \) is the freezing-point depression constant (for water, it is 1.86 \(^{\circ} \mathrm{C/molal}\)).

In the step-by-step solution, the freezing-point depression was calculated to be very small, approximately \(2.1 \times 10^{-5}\) °C. This minor change shows the limitation of using freezing-point depression for determining the molar mass of large molecules, as small changes could lead to inaccurate results. In such cases, larger magnitude measurements like osmotic pressure are preferable.

Osmotic Pressure

Osmotic pressure is another important colligative property that measures the tendency of a solvent to move across a semipermeable membrane. Osmotic pressure increases with higher solute concentration and provides more significant measurements, especially useful for large molecules.

• The formula for osmotic pressure is: \[ \pi = \frac{nRT}{V} \]
• Where \( n \) is the number of moles of solute, \( R \) is the ideal gas constant, \( T \) is the absolute temperature in Kelvin, and \( V \) is the volume of the solution in liters.

In the solution example, osmotic pressure was calculated to be 0.0027 atm, which is a more substantial value compared to the freezing-point depression. This makes osmotic pressure more suitable for determining the molar masses of large molecules because it allows for more precise and accurate measurements.

Molality

Molality is a concentration unit defined as the number of moles of solute per kilogram of solvent. It is particularly useful in colligative properties because it is not affected by temperature changes, unlike molarity.

• Molality is calculated using the formula: \[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

In the exercise, the molality of the protein was found to be \(1.11 \times 10^{-5}\) mol/kg. This value was used in the calculations of both freezing-point depression and osmotic pressure. Because molality depends only on the quantity of the solvent, it provides a reliable basis for measuring colligative properties even when solutions are subject to varying thermal conditions.

Molar Mass Determination

Determining the molar mass of a substance using colligative properties involves measuring changes in the physical properties of a solution. The choice between using freezing-point depression or osmotic pressure depends on the precision required and the molecular size of the solute.

• Freezing-point depression is less effective for large molecules due to the tiny temperature change, which can result in significant measurement errors.
• Osmotic pressure is generally more reliable for large molecules, providing larger and more distinguishable measurement values.

In the scenario given, osmotic pressure is recommended for determining the molar mass of large proteins. It avoids complications linked with phase changes. This method offers enhanced accuracy and is especially useful in biochemical applications involving large-molecule solutes.