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.

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

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

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.

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.