Question

A chemist dissolves \(10 \mathrm{~g}\) of an unknown protein in a liter of water at \(25^{\circ} \mathrm{C}\). The osmotic pressure is found to be \(9.25\) \(\mathrm{mmHg}\). What is the protein's molecular weight. Based upon the number of moles in \(10 \mathrm{~g}\) of protein, what would the freezing point depression and boiling point elevation be? Assume \(\mathrm{R}=\) Universal Gas Constant \(=.0821\) (liter\(\left.\mathrm{atm} / \mathrm{mole}^{\circ} \mathrm{K}\right), \mathrm{k}_{\mathrm{f}}=\left(1.86^{\circ} \mathrm{C} / \mathrm{m}\right)\), and \(\mathrm{k}_{\mathrm{b}}=\left(.52^{\circ} \mathrm{C} / \mathrm{m}\right)\)

Short answer

The molecular weight of the protein is approximately \(198190.13 \ g/mol\). The freezing point depression is approximately \(0.000939^{\circ} \mathrm{C}\), and the boiling point elevation is approximately \(0.000262^{\circ} \mathrm{C}\).

Step-by-step solution

1. Calculate the number of moles

We will use the osmotic pressure formula, which is given by: \( \Pi = n * R * T / V \), where: - \( \Pi \) is the osmotic pressure, - n is the number of moles of the solute, - R is the universal gas constant, - T is the temperature in Kelvin, and - V is the volume of the solution in liters. We are given: - the osmotic pressure as \(9.25 \ mmHg\), - the gas constant as \(0.0821 \ L \cdot atm / (mol \cdot K)\), - the temperature as \(25^{\circ} \mathrm{C}\), and - the volume as \(1 \ L\). First, let's convert the osmotic pressure to atmosphere: \( 9.25 \ mmHg * \frac{1 \ atm}{760 \ mmHg} = 0.01217 \ atm\), and the temperature to Kelvin: \( 25^{\circ} \mathrm{C} + 273.15 = 298.15 \ K\). Now, we can rearrange the osmotic pressure formula to solve for the number of moles: \(n=\frac{\Pi * V}{R * T}\). Substitute the given values: \(n=\frac{0.01217 \ atm * 1 \ L}{0.0821 \ L \cdot atm / (mol \cdot K) * 298.15 \ K}\).

2. Calculate the molecular weight

Calculate the number of moles: n = \(5.047 * 10^{-5} \ mol\) We're given that there are \(10 \ g\) of protein. In order to find the molecular weight (M) of the protein, we divide the mass (m) by the number of moles (n): \(M = \frac{m}{n} = \frac{10 \ g}{5.047 * 10^{-5} \ mol}\). Calculate the molecular weight (M): M = \(198190.13 \ g/mol\).

3. Calculate the molality

The molality (m) of a solution is the number of moles of solute per kilogram of solvent (in this case, water). Since there are 1,000 grams of water in a kilogram, the molality is: \(m=\frac{n}{1 \ kg}=\frac{5.047 * 10^{-5} \ mol}{1 \ kg}\). Calculate the molality: m = \(5.047 * 10^{-5} \ mol/kg\).

4. Calculate the freezing point depression and boiling point elevation

Now we're ready to find the freezing point depression and boiling point elevation. We're given the values of \(k_f\) and \(k_b\), which are the cryoscopic and ebullioscopic constants, respectively: - \(k_f = 1.86^{\circ} \mathrm{C} / m\), - \(k_b = 0.52^{\circ} \mathrm{C} / m\). The formulas for the freezing point depression (\(ΔT_f\)) and boiling point elevation (\(ΔT_b\)) are: \(\Delta T_f=k_f \cdot m\), \(\Delta T_b=k_b \cdot m\). Substitute the given values of molality (m) into the formulas: \(\Delta T_f=1.86^{\circ} \mathrm{C}/m * 5.047 * 10^{-5} \ mol/kg\) \(\Delta T_b=0.52^{\circ} \mathrm{C}/m * 5.047 * 10^{-5} \ mol/kg\). Calculate the freezing point depression and boiling point elevation: \(\Delta T_f \approx 0.000939^{\circ} \mathrm{C}\) \(\Delta T_b \approx 0.000262^{\circ} \mathrm{C}\). So, the freezing point depression is approximately \(0.000939^{\circ} \mathrm{C}\), and the boiling point elevation is approximately \(0.000262^{\circ} \mathrm{C}\).

Key concepts

Molecular Weight Calculation

The concept of molecular weight calculation is crucial when analyzing solutions, especially when determining the identity of unknown compounds. We start by calculating the number of moles of the solute using the osmotic pressure formula:- Formula: \( \Pi = \frac{nRT}{V} \).Here, \( \Pi \) is the osmotic pressure, \( n \) is the number of moles, \( R \) is the universal gas constant (0.0821 L·atm/mol·K), \( T \) is the temperature in Kelvin, and \( V \) is the volume in liters.By rearranging, we solve for \( n \), the number of moles: \( n = \frac{\Pi V}{RT} \). Once the moles are known, the molecular weight \( M \) can be calculated. The molecular weight is given by \( M = \frac{m}{n} \), where \( m \) is the mass of the substance in grams.In our example, using the protein mass of 10 g and calculated moles, the molecular weight turns out to be incredibly large, approximately 198,190.13 g/mol. This suggests a complex or large biomolecule, which is typical for proteins.

Freezing Point Depression

Freezing point depression occurs when a solute is added to a solvent, causing the solution to freeze at a lower temperature than the pure solvent. This is a colligative property, meaning it depends on the number of solute particles, not their identity.The formula for freezing point depression is:- \( \Delta T_f = k_f \cdot m \),where \( \Delta T_f \) is the decrease in freezing temperature, \( k_f \) is the cryoscopic constant of the solvent (1.86 °C/m for water), and \( m \) is the molality of the solution. Molality is the number of moles of solute per kilogram of solvent.In the given exercise, the molality is extremely low, around \( 5.047 \times 10^{-5} \ mol/kg \). Multiplying this with the cryoscopic constant gives a negligible freezing point depression of 0.000939°C, indicating the small effect tiny amounts of protein have on the freezing point of water.

Boiling Point Elevation

Similar to freezing point depression, boiling point elevation is a colligative property that results when a non-volatile solute raises the boiling point of a solvent.The relevant equation is:- \( \Delta T_b = k_b \cdot m \),where \( \Delta T_b \) is the increase in boiling temperature, \( k_b \) is the ebullioscopic constant for the solvent (0.52 °C/m for water), and \( m \) is the molality.For the protein solution in the exercise, we use the same low molality. The tiny resultant increase in boiling point (0.000262°C) shows that significant solute quantities are needed to noticeably change boiling points, which is often useful for understanding the behavior of solutions in laboratory and industrial settings.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, helping connect properties of gases under various conditions. It states:- \( PV = nRT \),where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin. This law applies to ideal gases, which perfectly follow the assumptions of kinetic molecular theory.While the ideal gas law typically describes gaseous systems, its components, such as the universal gas constant \( R \), appear in calculations involving solutions and osmotic pressure as well. When solving for osmotic pressure, we conceptually treat the solution as behaving somewhat like an ideal gas to determine the number of moles, illustrating the versatility of this principle across different states of matter.