Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g − 1 and 16 mg g − 1 aqueous phenol solutions, the concentrations of adsorbe

Question

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from $10 mg g^{- 1}$ and $16 mg g^{- 1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be $4 mg g^{- 1}$ and $10 mg g^{- 1}$ , respectively. At this temperature, the concentration (in $mg g^{- 1}$ ) of adsorbed phenol from $20 mg g^{- 1}$ aqueous solution of phenol will be .
Use : $lo g_{10} 2 = 0.3$

TestHub · Accepted Answer

​ m x ​ = K × C 1/ n lo g ( m x ​ ) = lo g K + n 1 ​ lo g C lo g 4 = lo g K + n 1 ​ lo g 10 0.6 = lo g K + n 1 ​ ..... ( 1 ) lo g 10 = lo g K + n 1 ​ lo g 16 1 = lo g K + n 1 ​ × 1.2.... ( 2 ) ​ Equation (2) - equation (1) 0.4 = n 1 ​ × ( 0.2 ) ⇒ n = 0.5 and lo g K = − 1.4 ​ lo g m x ​ = lo g K + n 1 ​ × lo g C = − 1.4 + 2 × lo g 20 ​ ​ = − 1.4 + 2.6 = 1.2 m x ​ = 1 0 + 1.2 = 16 ( lo g 2 = 0.3 , 4 lo g 2 = 1.2 , 16 = 1 0 + 1.2 ) ​ ​ m x ​ = K × C 1/ n 4 = K ( 10 ) 1/ n .... ( 1 ) 10 = K ( 16 ) 1/ n .... ( 2 ) X = K ( 20 ) 1/ n .... ( 3 ) ​ On solving equation (1) and (2) n 1 ​ = 2 On solving equation (1) and (3) X 4 ​ = ( 20 10 ​ ) 2 X = 16 On solving equation (2) and (3) ​ X 10 ​ = ( 20 16 ​ ) 2 X = 15.625 ​

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