Home
Class 12
CHEMISTRY
How many grams of gas would be adsobed p...

How many grams of gas would be adsobed per gram of a substance at `8` atm by assuming Freundlich adsorption isotherm.
`x/m=kp^(1//n)`
and `k=10^(-2) atm^(-1//3)" "&" "n=3`

Text Solution

Verified by Experts

The correct Answer is:
`x/m=10^(-2)xx8^(1//3)=0.02g`
Promotional Banner

Similar Questions

Explore conceptually related problems

For Freundlich adsorption isotherm, (x)/(m)=kp^1//n , the value of n is

How many moles are three in 1m^(3) of any gas at 0^(@)C and 1 atm :-

How many grams of NH_(4)NO_(3) should be dissolved per litre of solution to have a ph of 5.13 ? K_(b) for NH_(3) iss 1.8 xx10^(3)

Freundlich adsorption isotherm is obeyed by the adsorption where the adsorbate forms a multimolecular layer on the surface of adsorbent .In such cases, the degree of adsorption varies linearly with pressure but at high pressure, it becomes independent of pressure The relation of Freundlich adsorption isotherm is : (x)/(m)=kP^(1//n) where ,K and n are constants. Langmuir adsorption isotherm is obeyed by the adsorption where the adsorbate forms only a unimolecular adsorbed layer. The mathematical relation of Langmuir isotherm is : (x)/(m) = (aP)/(1+bP) In the mathematical relation of Freundlich adsorption isotherm ,the value of (1/n) is 0 le (1)/(n) le 1 . (a) True (b) False

Freundlich adsorption isotherm is obeyed by the adsorption where the adsorbate forms a multimolecular layer on the surface of adsorbent .In such cases, the degree of adsorption varies linearly with pressure but at high pressure, it becomes independent of pressure The relation of Freundlich adsorption isotherm is : (x)/(m)=kP^(1//n) where ,K and n are constants. Langmuir adsorption isotherm is obeyed by the adsorption where the adsorbate forms only a unimolecular adsorbed layer. The mathematical relation of Langmuir isotherm is : (x)/(m) = (aP)/(1+bP) Freundlich adsorption isotherm is valid for chemisorption . (a) True (b) False

Freundlich adsorption isotherm is obeyed by the adsorption where the adsorbate forms a multimolecular layer on the surface of adsorbent .In such cases, the degree of adsorption varies linearly with pressure but at high pressure, it becomes independent of pressure The relation of Freundlich adsorption isotherm is : (x)/(m)=kP^(1//n) where ,K and n are constants. Langmuir adsorption isotherm is obeyed by the adsorption where the adsorbate forms only a unimolecular adsorbed layer. The mathematical relation of Langmuir isotherm is : (x)/(m) = (aP)/(1+bP) When "log"((x)/(m)) is plotted against log P, we get a straight line with slope (1/n). (a) True (b) False

Freundlich adsorption isotherm is obeyed by the adsorption where the adsorbate forms a multimolecular layer on the surface of adsorbent .In such cases, the degree of adsorption varies linearly with pressure but at high pressure, it becomes independent of pressure The relation of Freundlich adsorption isotherm is : (x)/(m)=kP^(1//n) where ,K and n are constants. Langmuir adsorption isotherm is obeyed by the adsorption where the adsorbate forms only a unimolecular adsorbed layer. The mathematical relation of Langmuir isotherm is : (x)/(m) = (aP)/(1+bP) The degree of adsorption (x/m) at low pressure will be : (x)/(m) =a (a) True (b) False