The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is \underline\hspace2cm g mol ^-1 (Nearest integer) (Given : R = 0.083 L atm K ^-1 mol ^-1 )

Solutions PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q17 with Solution

View written solutionFree

Correct answer: 41500

For a dilute polymer solution,

$\pi = \frac{n_2}{V}RT = \frac{w_2}{M_2 V}RT$

Rearranging,

$\frac{\pi}{C} = \frac{RT}{M_2}$

where $C = \dfrac{w_2}{V}$ in g L $^{-1}$ .

So, from a graph of osmotic pressure $\pi$ versus concentration $C$ , the slope is

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

Hence,

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

From the plotted line, the ratio is approximately

$\text{slope} \approx 6.0 \times 10^{-4}\ \text{atm L g}^{-1}$

Given:

$R = 0.083\ \text{L atm K}^{-1}\text{mol}^{-1}, \qquad T = 300\ \text{K}$

So,

$RT = 0.083 \times 300 = 24.9\ \text{L atm mol}^{-1}$

Now,

$M = \frac{24.9}{6.0\times 10^{-4}}$

$M = 4.15\times 10^4\ \text{g mol}^{-1}$

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

The molar mass of PVC is

$\boxed{41500\ \text{g mol}^{-1}}$

This matches the stored correct answer.