Question

Plot of log against log P is a straight line inclined at an angle of `45^(@)`. 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 )
A. 1 g
B. 2 g
C. 3 g
D. 5 g

Answer 1

Correct Answer - D
According to Freundlich adsorption isotherm:
`"log"((x)/(m))=log k +(1)/(n)"log"P`
When `"log"((x)/(m))` is plotted against log P , we get straight line of slope (1/n) and intercept (log `k`).
`(1)/(n)=tan 45^(@)=1`
`"log" k ="log" 10=1`
`(x)/(m)=k(P)^(1//n)=10(0.5)^(1)=5`
When `m=1g , x=5g`