A 2.26 g sample of a protein of unknown molecular weight was dissolved in enough water to produce 21.3 mL of solution. The solution was found to have an osmotic pressure of 2.45 torr at 25 °C. What is the molecular weight of the protein?

• A. 100,000 g/mol
• B. 12,000 g/mol
• C. 68,000 g/mol
• D. 45,000 g/mol

Answer: (C)

Verified step by step guidance

1. Start by understanding the concept of osmotic pressure, which is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. It is related to the concentration of solute particles in the solution.
2. Use the formula for osmotic pressure: \( \Pi = \frac{nRT}{V} \), where \( \Pi \) is the osmotic pressure, \( n \) is the number of moles of solute, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the solution in liters.
3. Convert the osmotic pressure from torr to atm, since the ideal gas constant \( R \) is typically used in units of atm·L/mol·K. Use the conversion factor: \( 1 \text{ atm} = 760 \text{ torr} \).
4. Calculate the temperature in Kelvin by converting from Celsius using the formula: \( T(K) = T(°C) + 273.15 \).
5. Rearrange the osmotic pressure formula to solve for the number of moles \( n \): \( n = \frac{\Pi V}{RT} \). Then, use the definition of molecular weight \( MW = \frac{\text{mass}}{n} \) to find the molecular weight of the protein, where mass is given as 2.26 g.