Q. Plot of $\log{x/m}$ against log p is a straight line inclined at an angle of $45{^\circ}$. When the pressure is 0.5 atm and Freundlich parameter, k is 10, the amount of solute adsorbed per gram of adsorbent will be (log 5=0.6990)
A) 1 g
B) 2 g
C) 3 g
D) 5 g
Explanation
(d)
Freundlich adsorption isotherm equation is
$\frac{x}{m} = kp^{1/n}$
On taking log both sides
$\log{\frac{x}{m} = \log{k + \frac{1}{n}\log p}}$
$\log{\frac{x}{m} = \log{10 + \frac{1}{n}\log{0.5}}}$
($\because Slope = \frac{1}{n} = \tan\theta = \tan 45{^\circ} = 1)$
$\log{\frac{x}{m} = 1 + \frac{1}{1}\log\left( 5 \times 10^{- 1} \right)}$
$\log{\frac{x}{m} = 1 + 0.6990 - 1}$
=0.6990
$\frac{x}{m} = 5.00$
=5 g
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.