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← JEE Advanced 2025 · Paper 2
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JEE Advanced 2025 · Paper 2, Chemistry question 11

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g−110\ \mathrm{mg\,g^{-1}} and 16 mg g−116\ \mathrm{mg\,g^{-1}} aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g−14\ \mathrm{mg\,g^{-1}} and 10 mg g−110\ \mathrm{mg\,g^{-1}}, respectively. At this temperature, the concentration (in mg g−1\mathrm{mg\,g^{-1}}) of adsorbed phenol from 20 mg g−120\ \mathrm{mg\,g^{-1}} aqueous solution of phenol will be _____.

Use: log⁡102=0.3\log_{10}2 = 0.3

Answer and worked solution

Official answer: 15.5

Freundlich isotherm: xm=kC1/n\frac xm=kC^{1/n}.

Exponent. Divide the two data points:
104=(1610)1/n.\frac{10}{4}=\left(\frac{16}{10}\right)^{1/n}.
Taking logs, log⁡2.5=0.4\log2.5=0.4 and log⁡1.6=0.2\log1.6=0.2, so 1n=2\frac1n=2.

Constant. k=4102=0.04k=\frac{4}{10^2}=0.04.

Answer. At C=20C=20: xm=0.04×400=16\frac xm=0.04\times400=16 mg g⁻¹.