Plot of log left( frac{x}{m} right) against l | vedclass

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(D) According to the Freundlich adsorption isotherm equation: $\log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P$ Comparing this with the

Vedclass · Accepted Answer

$5$

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(D) According to the Freundlich adsorption isotherm equation: $\log \left( \frac{x}{m} ight) = \log k + \frac{1}{n} \log P$ Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$. Given that the line is inclined at an angle of $45^o$,the slope is: $\frac{1}{n} = 	an 45^o = 1$ Given $k = 10$ and $P = 0.5 \ atm$,we substitute these values into the adsorption isotherm equation: $\frac{x}{m} = k 	imes P^{1/n}$ $\frac{x}{m} = 10 	imes (0.5)^1$ $\frac{x}{m} = 5$ Since we are calculating the amount of solute adsorbed per gram of adsorbent $(m = 1 \ g)$,the amount of solute $x$ is $5 \ g$.

Plot of $\log \left( \frac{x}{m} \right)$ against $\log P$ is a straight line inclined at an angle of $45^o$. When the pressure is $0.5 \ atm$ and Freundlich parameter,$k$ is $10$,the amount of solute adsorbed per gram of adsorbent will be ...... $g$ $(\log 5 = 0.6990)$

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