QJEEM23Chac65: JEE Main 2023 Chemistry - Class 12 JEEM23Chac65 Solution: An aqueous solution of volume 300   cm 3 contains 0.63…

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:

$$\mathrm{\Pi }=\frac{n}{V}RT$$

where:

• $\Pi$ is the osmotic pressure,
• $n$ is the number of moles of solute,
• $V$ is the volume of the solution,
• $R$ is the ideal gas constant, and
• $T$ is the temperature in Kelvin.

First, we convert all given quantities to the appropriate units.

• $\Pi = 1.29$ mbar is equivalent to $1.29 \times 10^{-3}$ bar (since 1 bar = 1000 mbar),
• $V = 300$ cm³ is equivalent to $0.3$ L (since 1 L = 1000 cm³),
• $T = 300$ K,
• $R = 0.083$ L bar K $^{-1}$ mol $^{-1}$ .

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.

So the equation becomes:

$$\mathrm{\Pi }=\frac{m}{MV}RT$$

We are trying to solve for $M$ , so rearrange the equation to isolate $M$ :

$$M=\frac{mRT}{\mathrm{\Pi }V}$$

Finally, substitute the given values into the equation:

$$M=\frac{0.63\times 0.083\times 300}{1.29\times {10}^{-3}\times 0.3}\approx 40535\text{g/mol}$$

So the molar mass of the protein is approximately 40535 g/mol.