Applying Freundlich adsorptin isotherm, calculate the amount of acetic acid adsorbed by 1 kg of blood charcoal at 25^(@)C from a 5% vinegar solution (mass/volume). Given that if the concentration is expressed in molarity (mol dm^(-3)), x/m is mass of the solute adsorbed per gram of adsorbent, then k=0.160and n=2.32.

Applying Freundlich adsorptin isotherm, calculate the amount of acetic

1.	Applying Freundlich adsorptin isotherm, calculate the amount of acetic acid adsorbed by 1 kg of blood charcoal at 25^(@)C from a 5% vinegar solution (mass/volume). Given that if the concentration is expressed in molarity (mol dm^(-3)), x/m is mass of the solute adsorbed per gram of adsorbent, then k=0.160and n=2.32.
Answer» Solution :According to Freundlich adsorption isotherm, `x/m=k C^(1//n)""…(i)` `5%` vingar (acetic acid solution) means 5 G of acetic acid are present in 100 mL of the solution Molar mass of acetic acid `(CH_(3)COOH)=60 g mol^(-1)""therefore 5g` of acetic acid `=5/60` mol `therefore` 1000 mL of the solution will contain acetic acid `=5/60xx1/100xx1000=0.837` mol i.e., CONCENTRATION of the solution `(C ) =0.837 mol L ` Substtutin the values in eqn. (i), we GET `x/m=0.160xx(0.837)^(1//2.32)` `therefore log ""(x)/(m)=log(0.160)+(1)/(2.32)log (0.837)=-0.7959+0.435(-0.0773)=-0.8295=bar1.1750` `therefore x/m=` Antilog `(bar1.1750)=0.1481g (g CHARCOAL)""^(-1)` `therefore` Amount adsorbed by 1 kg (1000 g) of charcoal `=148.1g.`