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# Plot of $\;log \large\frac{x}{m}\;$ against $\;log P\;$ is a straight line inclined at an angle of $\;45^{0}\;$ . When the pressure is 0.5 atm and Freundlich parameter , K is 10.6 , the amount of the solute adsorbed per gram of adsorbed will be $\;(log 5 = 0.0990)\;$

$(a)\;1\;g\qquad(b)\;2\;g\qquad(c)\;3\;g\qquad(d)\;5\;g$

Answer : $\;1\;g$
Explanation :
According to freundlich equation , $\;\large\frac{x}{m}=KP^{\large\frac{1}{n}}\;$
or $\;log \large\frac{x}{m}=log K + \large\frac{1}{n} log P$
Therefore , plot of $\;log \large\frac{x}{m}\;$ vs log P is a straight line with slope = $\;\large\frac{1}{n}\;$ and intercept = log K
Thus , $\;\large\frac{1}{n}=tan \theta = tan 45^{0}=1\;or\;n=1$
At $\;P=0.5 atm\;$ and K=10
$\large\frac{x}{m}=10 (0.5)=5$
When m=1 g
Therefore , x (i . e , amount adsorbed per gram ) = 5 g