JEE Chemistry 2023 Question with Solution

Given: mass of solute = $2.5 \, \text{g}$, volume of solution = $250.0 \, \text{mL} = 0.250 \, \text{L}$, osmotic pressure = $400 \, \text{Pa}$, temperature = $300 \, \text{K}$.

Find: molar mass of the solute.

Use the osmotic pressure relation:

$$\pi = CRT$$

Now substitute concentration in terms of molar mass:

$$C = \frac{\frac{2.5}{M_o}}{\frac{250}{1000}}$$

Using the conversion $400 \, \text{Pa} = \frac{400}{10^5} \, \text{bar}$, we get:

$$\frac{400}{10^5} = \frac{\frac{2.5}{M_o}}{\frac{250}{1000}} \times 0.83 \times 300$$

Solving gives:

$$M_o = 62250$$

So, the answer is 62250.

Given: osmotic pressure formula, mass of solute, volume of solution, and temperature.

Find: molar mass of the solute.

The osmotic pressure of a solution is given by:

$$\pi = MRT$$

where $M$ is the molarity of the solution.

Molarity is:

$$\text{Molarity }(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$$

Convert volume:

$$\text{Volume of solution} = \frac{250.0 \, \text{mL}}{1000} = 0.250 \, \text{L}$$

Let the molar mass be $x \, \text{g/mol}$. Then:

$$\text{moles of solute} = \frac{2.5 \, \text{g}}{x \, \text{g/mol}}$$

Hence,

$$M = \frac{\frac{2.5}{x}}{0.250}$$ $$M = \frac{10}{x} \; \text{mol/L}$$

Now apply:

$$\pi = MRT$$ $$400 \, \text{Pa} = \frac{10}{x} \times 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} \times (27^\circ \text{C} + 273.15) \, \text{K}$$

The extracted detailed solution also states after simplifying:

$$x = 165 \; \text{g mol}^{-1}$$

but then concludes, "Therefore, the molar mass of the solute is $165 \, \text{g mol}^{-1}$, i.e., 62250." This is internally inconsistent. Since the solution prominently lists Correct Answer: 62250 and the first approach also concludes $M_o = 62250$, the extracted source supports the final answer as 62250, while noting a discrepancy in the second approach.