Problem 51 - Chemistry: Principles and Reactions

Question

A biochemist isolates a new protein and determines its molar mass by osmotic pressure measurements. A 50.0-mL solution is prepared by dissolving \(225 \mathrm{mg}\) of the protein in water. The solution has an osmotic pressure of \(4.18 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). What is the molar mass of the new protein?

Step-by-Step Solution

Answer

Answer: The molar mass of the protein is approximately 202 g/mol.

Step 1: Convert the given information to the appropriate units

First, we need to convert the given information to the appropriate units: 1. Volume: Given volume = \(50.0~\mathrm{mL}\), convert it into liters: 1 L = 1000 mL \(50.0~\mathrm{mL} \times \dfrac{1~\mathrm{L}}{1000~\mathrm{mL}} = 0.0500~\mathrm{L}\) 2. Osmotic pressure: Given pressure = \(4.18~\mathrm{mm~Hg}\), convert it into atm: 1 atm = 760 mm Hg \(4.18~\mathrm{mm~Hg} \times \dfrac{1~\mathrm{atm}}{760~\mathrm{mm~Hg}} = 0.00550~\mathrm{atm}\) 3. Temperature: Given temperature = \(25^{\circ}\mathrm{C}\), convert it into Kelvin: K = °C + 273.15 \(25 + 273.15 = 298.15~\mathrm{K}\)

Step 2: Calculate the moles of protein

Now, we use the osmotic pressure formula, \(\pi = \dfrac{nRT}{V}\), and solve for the moles of protein, \(n\): \(\pi = 0.00550~\mathrm{atm}\), \(R = 0.0821~\mathrm{\dfrac{L \cdot atm}{mol \cdot K}}\), \(T = 298.15~\mathrm{K}\), and \(V = 0.0500~\mathrm{L}\) \(n = \dfrac{\pi V}{RT} = \dfrac{(0.00550~\mathrm{atm})(0.0500~\mathrm{L})}{(0.0821~\mathrm{\dfrac{L \cdot atm}{mol \cdot K}})(298.15~\mathrm{K})} = 1.115 \times 10^{-3}~\mathrm{mol}\)

Step 3: Calculate the molar mass of the protein

Next, we will use the moles of protein and the given mass of protein to find the molar mass: Given mass of protein = \(225~\mathrm{mg}\), convert it into grams: 1 g = 1000 mg \(225~\mathrm{mg} \times \dfrac{1~\mathrm{g}}{1000~\mathrm{mg}} = 0.225~\mathrm{g}\) Molar mass = \(\dfrac{\text{mass of protein}}{\text{moles of protein}} = \dfrac{0.225~\mathrm{g}}{1.115 \times 10^{-3}~\mathrm{mol}} = 202~\mathrm{\dfrac{g}{mol}}\) The molar mass of the new protein is approximately \(202~\mathrm{\dfrac{g}{mol}}\).

Key Concepts

Osmotic Pressure Calculations

Osmotic pressure is a fundamental concept often used in chemistry and biology to determine the concentration of a solute in a solution. It can be described as the pressure required to halt the osmotic flow of water through a semipermeable membrane separating two solutions with different concentrations.

The osmotic pressure, \( \pi \), can be calculated using the formula \( \pi = \frac{nRT}{V} \), where \( n \) represents the number of moles of solute, \( R \) the gas constant (\(0.0821~\frac{L \cdot atm}{mol \cdot K}\)), \( T \) the temperature in Kelvins, and \( V \) the volume of the solution in liters.

For our example with the protein, the osmotic pressure was given, and we wanted to find the molar mass. We first calculated the number of moles of the protein from the osmotic pressure, then used the mass of the dissolved protein to determine its molar mass. Remember, accurate determination of osmotic pressure is crucial for calculating the molar mass of solutes accurately, especially in biochemical applications.

Unit Conversion in Chemistry

Unit conversion is a vital skill in chemistry that ensures measurements are standardized and comparisons between different experiments are possible. When working with any formula or calculation in chemistry, it is crucial to convert all the units to the appropriate system (often the SI system) to ensure accuracy in the results.

For instance, in the protein molar mass problem, we converted the volume from milliliters (mL) to liters (L), the pressure from millimeters of mercury (mm Hg) to atmospheres (atm), and the temperature from Celsius to Kelvin. We must also convert the mass from milligrams (mg) to grams (g) before using it in the molar mass calculation. Paying close attention to these conversions helps avoid errors in the final results.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the ratio of the number of solute particles to the solvent, regardless of the solute's identity. Osmotic pressure is one of these properties, along with boiling point elevation, freezing point depression, and vapor pressure lowering.

Understanding colligative properties is crucial for solving problems related to solution concentration. For example, in our osmotic pressure scenario, the molar mass of the protein is derived by relating the osmotic pressure (a colligative property) to the number of moles of solute in the solution.

Therefore, recognizing that osmotic pressure does not depend on the type of protein but on the number of protein molecules in the solution is key to finding the correct molar mass. This understanding of colligative properties not only applies to pure chemistry but has real-world implications in food science, medicine, and environmental science.