Question

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 100, the amount of the solute adsorbed per gram of adsorbent will be (log 5 = 0.6990)

• A. 1 g
• B. 2 g
• C. 3 g
• D. 5 g

Answer 1

The correct option is D 5 g

According to Freundlich equation, $\frac{x}{m}=K{P}^{\frac{1}{n}}$

or $log\frac{x}{m}=logK+\frac{1}{nlogP}$

$\therefore$ Plot of log $\frac{x}{m}$ vs log P is a straight line

with slope = $\frac{1}{n}$ and intercept = log K. Thus,

$\frac{1}{n}$ = tan $\theta$ = tan ${45}^{\circ }$ = 1 or n = 1

At P = 0.5 atm and K = 10, $\frac{x}{m}$ = 10 $\times$ (0.5) = 5

$\therefore$ x (i.e. amount adsorbed per gram) = 5 g.