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 determine the molality of the solution. The molality (m) is given by the formula: \(m = \frac{moles\:of\:solute}{kilograms\:of\:solvent}\) Since we know the mass concentration of protein (\(1.0 g/L\)) and the molar mass of protein (\(9.0 \times 10^{4} g/mol\)), we can find the moles of solute by dividing mass concentration by the molar mass: \(moles\:of\:solute = \frac{1.0 g}{9.0 \times 10^{4} g/mol} = 1.11 \times 10^{-5} mol\) We also know the density of the solution is \(1.0 g/cm^3\) or \(1000 g/L\). Since the mass concentration of protein is low, we can assume the solvent mass to be almost equal to the solution mass. Thus, the mass of solvent is approximately \(1000 g (\approx 1 kg)\). Now we can calculate the molality: \(m = \frac{1.11 \times 10^{-5} mol}{1 kg} = 1.11 \times 10^{-5} mol/kg\)

The freezing-point depression is given by the formula: \(\Delta T_{f} = K_{f} \times m\) Where \(\Delta T_{f}\) is the freezing-point depression, \(K_{f}\) is the cryoscopic constant for water (which is \(1.86 ^{\circ}C/mol/kg\)), and m is the molality of the solution. We can plug in the molality we found in step 1: \(\Delta T_{f} = 1.86 ^{\circ}C/mol/kg \times 1.11 \times 10^{-5} mol/kg = 2.06 \times 10^{-5}^{\circ}C\)

The osmotic pressure is given by the formula: \(\Pi = n_i \times R \times T\) Where \(\Pi\) is the osmotic pressure, \(n_i\) is the moles of solute per liter of solution, R is the gas constant (which is \(0.0821 L \times atm/mol \times K\)), and T is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin: \(T = 25^{\circ}C + 273.15 = 298.15 K\) Now, we can calculate the moles of solute per liter of solution: \(n_i = 1.11 \times 10^{-5} mol/L\) Finally, we can find the osmotic pressure: \(\Pi = 1.11 \times 10^{-5} mol/L \times 0.0821 L \times atm/mol \times K \times 298.15 K = 2.71 \times 10^{-4} atm\)

A larger value of the colligative property would give better accuracy during molar mass measurements. Comparing the answers from Step 2 and Step 3, we find that the osmotic pressure (\(2.71 \times 10^{-4} atm\)) is larger than the freezing-point depression (\(2.06 \times 10^{-5}^{\circ}C\)). Therefore, using osmotic pressure would be a better method to determine the molar masses of large molecules. This is because osmotic pressure results in a larger value, leading to more precise measurements and less error.