Question

200 mL of an aqueous solution of a protein contains 1.26 g. The osmotic pressure of this solution at 300 K is found to be \( 2.57 \times 10^{-3} \, {bar} \). The molar mass of the protein will be: (R = 0.083 L bar mol\(^{-1}\) K\(^{-1}\))

• A. 51022 g/mol
• B. 122044 g/mol
• C. 31011 g/mol
• D. 61038 g/mol

Hint:

Osmotic pressure can be used to calculate the molar mass of a solute. Rearrange the formula to find the number of moles and then calculate the molar mass.

Answer 1

The Correct Option is D

Step 1: Osmotic pressure is given by the formula: \[ \Pi = \frac{nRT}{V}, \] where \( \Pi \) is the osmotic pressure, \( n \) is the number of moles of solute, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the solution.
Step 2: Rearranging the formula to solve for \( n \), the number of moles of solute: \[ n = \frac{\Pi V}{RT}. \] Substitute the given values: \[ n = \frac{(2.57 \times 10^{-3} \, {bar}) (0.200 \, {L})}{(0.083 \, {L bar mol}^{-1} {K}^{-1}) (300 \, {K})}. \] Step 3: Calculating the number of moles: \[ n = \frac{(2.57 \times 10^{-3})(0.200)}{(0.083)(300)} = 0.000206 \, {mol}. \] Step 4: The molar mass \( M \) of the protein is given by: \[ M = \frac{{mass of solute}}{n}. \] Substitute the values: \[ M = \frac{1.26 \, {g}}{0.000206 \, {mol}} = 61038 \, {g/mol}. \]