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: $$M = \frac{1.26 \, {g}}{0.000206 \, {mol}} = 61038 \, {g/mol}.$$