200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.

To determine the molar mass of the protein in the solution, we can use the formula for osmotic pressure, which is given by the equation:

Understanding Osmotic Pressure

The osmotic pressure (\( \Pi \)) of a solution can be expressed using the formula:

\(\Pi = \frac{n}{V}RT\)

Where:

• \( \Pi \) = osmotic pressure (in bar)
• \( n \) = number of moles of solute (in moles)
• \( V \) = volume of the solution (in liters)
• R = universal gas constant (0.0831 L·bar/(K·mol))
• T = temperature (in Kelvin)

Step-by-Step Calculation

We need to rearrange the formula to find the number of moles (\( n \)) of the protein:

\( n = \frac{\Pi V}{RT} \)

Now, let's plug in the values:

• \( \Pi = 2.57 \times 10^{-3} \, \text{bar} \)
• \( V = 200 \, \text{cm}^3 = 0.200 \, \text{L} \)
• \( R = 0.0831 \, \text{L·bar/(K·mol)} \)
• \( T = 300 \, \text{K} \)

Substituting these values into the equation:

\( n = \frac{(2.57 \times 10^{-3} \, \text{bar}) \times (0.200 \, \text{L})}{(0.0831 \, \text{L·bar/(K·mol)}) \times (300 \, \text{K})} \)

Calculating the denominator:

\( 0.0831 \times 300 = 24.93 \, \text{L·bar/mol} \)

Now, substituting this back into the equation for \( n \):

\( n = \frac{(2.57 \times 10^{-3}) \times (0.200)}{24.93} \)

Calculating the numerator:

\( 2.57 \times 10^{-3} \times 0.200 = 5.14 \times 10^{-4} \, \text{bar·L} \)

Now, we can find \( n \):

\( n = \frac{5.14 \times 10^{-4}}{24.93} \approx 2.06 \times 10^{-5} \, \text{mol} \)

Finding Molar Mass

The molar mass (\( M \)) of the protein can be calculated using the formula:

\( M = \frac{m}{n} \)

Where:

• \( m \) = mass of the protein (in grams)
• \( n \) = number of moles of the protein (in moles)

We know that the mass of the protein is 1.26 g. Now we can substitute the values:

\( M = \frac{1.26 \, \text{g}}{2.06 \times 10^{-5} \, \text{mol}} \)

Calculating this gives:

\( M \approx 61293.69 \, \text{g/mol} \)

Final Result

Thus, the molar mass of the protein in the solution is approximately 61294 g/mol. This value indicates the size of the protein molecules in the solution, which can be quite significant, reflecting the complexity of protein structures.