Plot of log `x/m` 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 `10.0` , the amount of the solute adsorbed per gram of adsorbent will be `(log 5=0.6990)` :

Options

A. 1 g
B. 2 g
C. 3 g
D. 5 g

Answer

D

Explanation

Freundlich adsorption isotherm equation is
`(x)/(m) = k_(p)^(1// n)`
On taking log both sides
`"log"(x)/(m) = log k + (1)/(n)` log p
`"log"(x)/(m) = log 10 + 1* log 0.5`
`"log" (x)/(m) = 1 + log (5 xx 10^(-1))`
`"log" (x)/(m) = 1 + 0.6990^(-1)`
= `0.6990`
`(x)/(m) = 5.00 = 5 ` g