QJEEM20Chbc6e: JEE Main 2020 Chemistry - Class 12 JEEM20Chbc6e Solution: For Freundlich adsorption isotherm, a plot of log (x/m)…

According to Freundlich adsorption isotherm, in the median range of pressure

       $$\frac{x}{m}\propto P^{\frac{1}{n}}$$

$$\Rightarrow$$  $$\frac{x}{m}$$ = kP$${}^{\frac{1}{n}}$$

taking log both sides, we get,

log$$\frac{x}{m}$$ = logk + $$\frac{1}{n}$$ logP

Here in graph between log$$\frac{x}{m}$$ and logP, slope is $$\frac{1}{n}$$ and intercepts = log k.

$$\therefore$$ $$\frac{1}{n}$$ = 2 $$\Rightarrow$$ n = $$\frac{1}{2}$$

and log k = 0.4771 = log 3

$$\Rightarrow$$ k = 3

$$\therefore$$ $$\frac{x}{m}$$ = 3P$${}^2$$

So, mass of gas adsorbed per gram of adsorbent

= 3 $$\times$$ (0.04)²

= 48 $$\times$$ 10^–4