Problem 73 - Chemistry An Atoms First Approach

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.

Step-by-Step Solution

Answer

The freezing-point depression is \(2.06 \times 10^{-5}^{\circ}C\), and the osmotic pressure is \(2.71 \times 10^{-4} atm\). Osmotic pressure would be a better method to determine the molar masses of large molecules because it results in a larger value, leading to more precise measurements and less error.

Step 1: Calculate the molality of the solution

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: \(m = \frac{1.11 \times 10^{-5} mol}{1 kg} = 1.11 \times 10^{-5} mol/kg\)

Step 2: Calculate the freezing-point depression

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: \(\Delta T_{f} = 1.86 ^{\circ}C/mol/kg \times 1.11 \times 10^{-5} mol/kg = 2.06 \times 10^{-5}^{\circ}C\)

Step 3: Calculate osmotic pressure

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: \(\Pi = 1.11 \times 10^{-5} mol/L \times 0.0821 L \times atm/mol \times K \times 298.15 K = 2.71 \times 10^{-4} atm\)

Step 4: Comparing the two colligative properties

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.

Key Concepts

Freezing-Point Depression

Freezing-point depression is a colligative property that describes the lowering of a liquid’s freezing point when a solute is added. It depends only on the number of dissolved particles, not on their identity. The degree of depression can be calculated using the formula: \[ \Delta T_f = K_f \times m \] where \( \Delta T_f \) is the freezing-point depression, \( K_f \) is the cryoscopic constant (a property of the solvent), and \( m \) is the molality of the solution. In the provided exercise, the freezing-point depression is quite small. This illustrates how solutions containing large molecules, like proteins, often have minimal impact on freezing-point changes. The minuscule drop, \( 2.06 \times 10^{-5} ^{\circ}C \), emphasizes the challenge of using this property for determining molar masses of large molecules with precision.
While freezing-point depression can be useful for smaller molar masses, its smaller measurable change makes it less preferable for larger molecules, which are better assessed using other colligative properties.

Osmotic Pressure

Osmotic pressure is another crucial colligative property, primarily dependable on the number of solute particles. It refers to the pressure required to prevent the flow of solvent into the solution through a semipermeable membrane. The formula for osmotic pressure is: \[ \Pi = n_i \times R \times T \] where \( \Pi \) is the osmotic pressure, \( n_i \) is the molarity expressed in moles per liter, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature (in Kelvin).
In the context of the provided problem, osmotic pressure yields a more considerable value, \( 2.71 \times 10^{-4} \text{ atm} \), compared to freezing-point depression. This makes it a potent method for determining molar masses of large molecules, where even slight measurement errors can significantly impact research outcomes.

• Osmotic pressure tends to offer more noticeable results for detecting solute concentrations.
• This makes it more reliable when dealing with complex data that often accompanies large molecule studies.

Molar Mass Determination

Determining the molar mass of a solute is vital for understanding its chemical and physical properties. Colligative properties like freezing-point depression and osmotic pressure are practical tools for these measurements. However, the choice of property largely depends on the solute's nature:
1. **Smaller Solutes:** Lower changes like freezing-point depression may suffice, offering basic insights without requiring sophisticated equipment.
2. **Larger Molecules:** As demonstrated in the exercise, osmotic pressure is more effective. By producing more noticeable and measurable changes, it's particularly useful for molecules with higher molar masses such as proteins. In practice, osmotic pressure facilitates more precise determinations by minimizing potential errors associated with lesser-detectable colligative properties.
The provided problem clearly shows why understanding the nature of solutes and colligative properties is essential when attempting to determine molar mass accurately.