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Poonam Rao
Asked: 2022-11-02T15:47:53+05:30 In:

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`

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`
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  1. 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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