Plot of log (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, the amount of solute
adsorbed per gram of adsorbent will be (log 5 : 0.6990):

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1 g

2 g

3 g

5 g

answer is D.

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Detailed Solution

According to Freundlich adsorption isotherm:

$\log (\frac{x}{m}) = \log k + \frac{1}{n} \log P$

When log (xlm) is plotted against Log P, we get straight line of slope (lln) and intercept (log k).

$\begin{array}{l} \frac{1}{n} = \tan 45^{\circ} = 1 \\ \log k = \log 10 = 1 \\ \frac{x}{m} = k (P)^{1 / n} = 10 (0 .5)^{1} = 5 \end{array}$

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