1.

Plot of log `x//m` against log `p` is a straigh line inclined at an angle of `46^(@)` . When the pressure is `0.5` atom and Freundlich parameter `k` is `10.0` , the amount of the solute adsorbed per gram of adsorbent will be `(log 5=0.6990)` :A. `2g`B. `1g`C. `5g`D. `3g`

Answer» Correct Answer - C
According to Freundlich adsorption isotherm
`(x)/(m)=kp^(1//m)`
or log `(x)/(m)=logk+(1)/(n)logP`
Plot of log `x//m` vs log `P` is a straight line with slope `=1//m` and intercept `=logk`. Thus
`(1)/(n)=tantheta=tan45^(@)=1`
or `n=1`
At `P=0.5` atm and `k=10`
`(x)/(m)=(10)(0.5)^(1)=5`
Thus, amount adsorbed per gram of adsorbent is `5g`


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