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 $(log\text{ }5=0.6990)$

• A. $1 \,g$
• B. $2\, g$
• C. $3\, g$
• D. $5\, g$

Answer 1

Answer: (D)

$5\, g$

Answer 2

Freundlich adsorption isotherm equation is
$\frac{x}{m}=k{{p}^{1/n}}$
On taking log both sides
$\log \frac{x}{m}=\log k+\frac{1}{n}\log \,p$
$\log \frac{x}{m}=\log \,10+\frac{1}{n}\log \,0.5$
( $\because$ $slope=\frac{1}{n}=tan\,\theta =tan\text{ }45{}^\circ =1$ ) $\log \frac{x}{m}=1+\frac{1}{1}\log (5\times {{10}^{-1}})$
$\log \frac{x}{m}=1-0.6990+1$ $=0.6990$
$\frac{x}{m}=5.00$
$=5g.$