An aqueous solution of volume $300 \mathrm{~cm}^{3}$ contains $0.63 \mathrm{~g}$ of protein. The osmotic pressure of the solution at $300 \mathrm{~K}$ is 1.29 mbar. The molar mass of the protein is ___________ $\mathrm{g} ~\mathrm{mol}^{-1}$

Given : R = 0.083 L bar K$^{-1}$ mol$^{-1}$

<p>The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal gas law:</p> <p>$\Pi = \frac{n}{V}RT$</p> <p>where:</p> <ul> <li>$\Pi$ is the osmotic pressure,</li> <li>$n$ is the number of moles of solute,</li> <li>$V$ is the volume of the solution,</li> <li>$R$ is the ideal gas constant, and</li> <li>$T$ is the temperature in Kelvin.</li> </ul> <p>First, we convert all given quantities to the appropriate units.</p> <ul> <li>$\Pi = 1.29$ mbar is equivalent to $1.29 \times 10^{-3}$ bar (since 1 bar = 1000 mbar),</li> <li>$V = 300$ cm³ is equivalent to $0.3$ L (since 1 L = 1000 cm³),</li> <li>$T = 300$ K,</li> <li>$R = 0.083$ L bar K$^{-1}$ mol$^{-1}$.</li> </ul> <p>Instead of the number of moles, we are given the mass of the solute. However, we can replace the moles with mass using the relationship $n = \frac{m}{M}$, where $M$ is the molar mass, and $m$ is the mass of the solute.</p> <p>So the equation becomes:</p> <p>$\Pi = \frac{m}{MV}RT$</p> <p>We are trying to solve for $M$, so rearrange the equation to isolate $M$:</p> <p>$M = \frac{mRT}{\Pi V}$</p> <p>Finally, substitute the given values into the equation:</p> <p>$$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3} \approx 40535 \, \text{g/mol}$$</p> <p>So the molar mass of the protein is approximately 40535 g/mol.</p>