Question

The plot of$\log x/m$ against $\log p$ is a straight line inclined at an angle of ${45}^{\circ }$. When the pressure is $0.5atm$ and Freundlich parameter, $k$ is $10$, the amount of solute adsorbed per gram of adsorbent will be: $(\log 5=0.6990)$

• A. $5g$
• B. $3g$
• C. $6g$
• D. $12g$

Answer 1

Answer: (A)

$5g$

The Freundlich adsorption isotherm is mathematically expressed as:

$\frac{x}{m}=K{P}^{\frac{1}{n}}$

It can be written in $y=mx+c$

$\log \frac{x}{m}=\frac{1}{n}\times \log P+\log K$

$\log \frac{x}{m}=\frac{1}{n}\log \left(0.5\right)+\log \left(10\right)$

Therefore, plot of $log\frac{x}{m}$ vs $logP$ is a straight line with slope = $\frac{1}{n}$ and intercept = log K

Thus ,$\frac{1}{n}=tan\theta =tan{45}^0=1$

$n=1$

$\log \frac{x}{m}=\log 5-\log \left(10\right)+\log \left(10\right)$

$\log \frac{x}{m}=\log 5$

$\frac{x}{m}=5$

Hence amount of solute per gram is $5g$.