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Q.

Graph between log x/m and log p is a straight line inclined at an angle of $45^{o}$ . When pressure is 0.5 atm and In k = 0.693, the amount of solute adsorbed per gram of adsorbent will be:

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a

0.25

b

1

c

1.5

d

2.5

answer is A.

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Detailed Solution

$\begin{array}{l} \log \frac{x}{m} = \log k + \frac{1}{n} \log p \\ \frac{1}{n} = \tan 45^{o}, ln k = 0 .693 \\ n = 1 \log k = 0 .3 \\ \log \frac{x}{m} = 0 .3 + \log (0 .5) \\ \log \frac{x}{m} = 0 .3 - 0 .3 = 0 \\ \frac{x}{m} = 10^{o} = 1 \end{array}$

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