Question:
Plot of $ log\text{ }x/m $ against log $p$ is a straight line inclined at an angle of $ 45{}^\circ $ . When the pressure is $ 0.5 \,arm $ and Freundlich parameter. $ k $ is $ 10 $ , the amount of solute adsorbed per gram of adsorbent will be $ (log\text{ }5=0.6990) $
Plot of $ log\text{ }x/m $ against log $p$ is a straight line inclined at an angle of $ 45{}^\circ $ . When the pressure is $ 0.5 \,arm $ and Freundlich parameter. $ k $ is $ 10 $ , the amount of solute adsorbed per gram of adsorbent will be $ (log\text{ }5=0.6990) $
Updated On: Jun 3, 2024
- $1 \,g$
- $2\, g$
- $3\, g$
- $5\, g$
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The Correct Option is D
Solution and Explanation
Freundlich adsorption isotherm equation is
$ \frac{x}{m}=k{{p}^{1/n}} $
On taking log both sides
$ \log \frac{x}{m}=\log k+\frac{1}{n}\log \,p $
$ \log \frac{x}{m}=\log \,10+\frac{1}{n}\log \,0.5 $
( $ \because $ $ slope=\frac{1}{n}=tan\,\theta =tan\text{ }45{}^\circ =1 $ ) $ \log \frac{x}{m}=1+\frac{1}{1}\log (5\times {{10}^{-1}}) $
$ \log \frac{x}{m}=1-0.6990+1 $ $ =0.6990 $
$ \frac{x}{m}=5.00 $
$ =5g. $
$ \frac{x}{m}=k{{p}^{1/n}} $
On taking log both sides
$ \log \frac{x}{m}=\log k+\frac{1}{n}\log \,p $
$ \log \frac{x}{m}=\log \,10+\frac{1}{n}\log \,0.5 $
( $ \because $ $ slope=\frac{1}{n}=tan\,\theta =tan\text{ }45{}^\circ =1 $ ) $ \log \frac{x}{m}=1+\frac{1}{1}\log (5\times {{10}^{-1}}) $
$ \log \frac{x}{m}=1-0.6990+1 $ $ =0.6990 $
$ \frac{x}{m}=5.00 $
$ =5g. $
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