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  4. Graph between log ((x/m)) and log P is straight line at angle of 45° with the intercept of 0.6020. <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/c-mettczvzadakbdfv.jpg /> The extent of adsorption ((x/m)) at a pressure of 1 atm is

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Q. Graph between $log \left(\frac{x}{m}\right)$ and log P is straight line at angle of $45^\circ $ with the intercept of 0.6020.
Question
The extent of adsorption $\left(\frac{x}{m}\right)$ at a pressure of 1 atm is

NTA AbhyasNTA Abhyas 2022

Solution:

$log \left(\frac{x}{m}\right)=log ⁡ K+\frac{1}{n}log ⁡ P$
$=0.602+1log 1$
= 0.602
So $\frac{x}{m}=4$ .