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Fruendlich Isotherm

Plot of log...

Question

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

[ $l o g 5 = 0.6990$ ]

1 g

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2 g

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3 g

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5 g

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Solution

The correct option is D $5 g$
According to Freundlich adsorption isotherm :

$l o g (\frac{x}{m}) = l o g k + \frac{1}{n} l o g P$

When log $l o g (\frac{x}{m})$ is plotted against log P, we get straight line of slope (1/n) and intercept (log k).

$\frac{1}{n} = t a n 45^{o} = 1$

$l o g k = l o g 10 = 1$

$\frac{x}{m} = k (P)^{1 / n} = 10 (0.5)^{1} = 5$

When $m = 1 g, x = 5 g$

Hence, option $D$ is correct.

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Plot of log(xm) against log P is a straight line inclined at an angle of 45o. When the pressure is 0.5 atm and Freundlich parameter, k is 10, the amount of solute adsorbed per gram of adsorbent will be: [log 5=0.6990]

The correct option is D 5 gAccording to Freundlich adsorption isotherm : log(xm)=logk+1nlogPWhen log log(xm) is plotted against log P, we get straight line of slope (1/n) and intercept (log k).1n=tan45o=1logk=log10=1xm=k(P)1/n=10(0.5)1=5 When m=1 g,x=5 gHence, option D is correct.

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

[ $l o g 5 = 0.6990$ ]