JEE Chemistry 2023 Question with Solution

Given: osmotic pressure data for PVC in cyclohexanone at $300 \, \text{K}$, and $R = 0.083 \, \text{L atm K}^{-1} \text{ mol}^{-1}$.

Find: the molar mass of PVC.

For polymer solutions, osmotic pressure is related by

$$\pi = C\left(\frac{RT}{M}\right)$$

Hence,

$$\frac{\pi}{C} = \frac{RT}{M}$$

So the graph of $\pi$ versus $C$ has slope

$$\text{slope} = \frac{RT}{M}$$

From the graph, the slope is

$$6 \times 10^{-4}$$

Therefore,

$$M = \frac{RT}{\text{slope}} = \frac{0.083 \times 300}{6 \times 10^{-4}}$$

Now,

$$M = 41500 \, \text{g mol}^{-1}$$

Therefore, the molar mass of PVC is $41500 \, \text{g mol}^{-1}$.

Given: the plotted graph is used to extract the proportionality between osmotic pressure and concentration.

Find: molar mass $M$ of PVC.

Starting with van't Hoff relation for dilute polymer solution,

$$\pi = C\left(\frac{RT}{M}\right)$$

Rearranging,

$$\frac{C}{\pi} = \frac{M}{RT}$$

Thus $\frac{C}{\pi}$ is a function of $C$, and equivalently for the straight-line interpretation used in the solution, the slope of the $\pi$ versus $C$ graph is

$$\frac{RT}{M}$$

Using the slope read from the graph,

$$\frac{RT}{M} = 6 \times 10^{-4}$$

So,

$$M = \frac{0.083 \times 300}{6 \times 10^{-4}} = 41500 \, \text{g mol}^{-1}$$

Hence, the required nearest integer is 41500.