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

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(D) According to the Freundlich adsorption isotherm,$\frac{x}{m} = K P^{1/n}$.Taking logarithm on both sides,we get $\log \left( \frac{x}{m} \right) =

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$5$

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(D) According to the Freundlich adsorption isotherm,$\frac{x}{m} = K P^{1/n}$.Taking logarithm on both sides,we get $\log \left( \frac{x}{m} ight) = \log K + \frac{1}{n} \log P$.The plot of $\log \left( \frac{x}{m} ight)$ versus $\log P$ is a straight line with slope $= \frac{1}{n}$ and intercept $= \log K$.Given that the line is inclined at an angle of $45^{\circ}$,the slope is $\frac{1}{n} = 	an 45^{\circ} = 1$.Thus,$n = 1$.Given $P = 0.5 \ atm$ and $K = 10$,substituting these values into the equation:$\frac{x}{m} = K 	imes P^{1/n} = 10 	imes (0.5)^{1} = 5$.Therefore,the amount of solute adsorbed per gram of adsorbent is $5 \ g$.

Plot of $\log \left( \frac{x}{m} \right)$ 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 $10$,then the amount of solute adsorbed per gram of adsorbent is .......... $g$ $(\log 5 = 0.699)$.

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