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JEE Advanced Chemistry Surface Chemistry 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Chemistry Question (2025) — Solution

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 : _ 10 2=0.3

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

aligned & x m =K C^ 1 / n \\ & ( x m )= K+ 1 n C \\ & 4= K+ 1 n 10 \\ & 0.6= K+ 1 n .....(1) \\ & 10= K+ 1 n 16 \\ & 1= K+ 1 n 1.2 ....(2) aligned Equation (2) - equation (1) 0.4= 1 n (0.2) n =0.5 and K =-1.4 aligned & x m = K+ 1 n C \\ & =-1.4+2 20 aligned aligned & =-1.4+2.6=1.2 \\ & x m =10^ +1.2 =16 \\ & ( 2=0.3,4 2=1.2,16=10^ +1.2 ) aligned aligned & x m =K C^ 1 / n \\ & 4=K(10)^ 1 / n ....(1) \\ & 10=K(16)^ 1 / n ....(2) \\ & X=K(20)^ 1 / n ....(3) aligned On solving equation (1) and (2) 1 n =2 On solving equation (1) and (3) 4 X = ( 10 20 )^2 X=16 On solving equation (2) and (3) aligned & 10 X = ( 16 20 )^2 \\ & X=15.625 aligned

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