JEE Advanced 2025 Paper 2 · Q11 · Rate Laws & Order
Adsorption of phenol from its aqueous solution onto fly ash obeys the Freundlich isotherm. At a given temperature, from 10 mg/g and 16 mg/g aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg/g and 10 mg/g, respectively. At this temperature, the concentration (in mg/g) of adsorbed phenol from a 20 mg/g aqueous solution of phenol will be __________.
Use: $\log_{10}(2) = 0.3$.
Reveal answer + step-by-step solution
Correct answer:16
Solution
Freundlich isotherm: $x/m = k \cdot C^{1/n}$. From the two data points, $10/4 = (16/10)^{1/n}$ -> $2.5 = 1.6^{1/n}$. Take log: $\log 2.5 = (1/n) \log 1.6$. $\log 2.5 = \log(10/4) = 1 - 2 \log 2 = 1 - 0.6 = 0.4$. $\log 1.6 = \log(16/10) = 4 \log 2 - 1 = 1.2 - 1 = 0.2$. $1/n = 0.4/0.2 = 2$. So $x/m = k C^2$. From $C = 10$, $x/m = 4$: $k = 4/100 = 0.04$. At $C = 20$: $x/m = 0.04 \times 400 = 16$ mg/g.
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