Chemistry · Surface Chemistry

JEE Advanced 2025 — Paper 2 — Question 40

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from $10 \mathrm{mg} \mathrm{g}^{-1}$ and $16 \mathrm{mg} \mathrm{g}^{-1}$ aqueous phenol solutions,

the concentrations of adsorbed phenol are measured to be $4 \mathrm{mg} \mathrm{g}^{-1}$ and $10 \mathrm{mg} \mathrm{g}^{-1}$ , respectively. At this temperature, the concentration (in $\mathrm{mg} \mathrm{g}^{-1}$ ) of adsorbed phenol from $20 \mathrm{mg} \mathrm{g}^{-1}$ aqueous solution of phenol will be $\qquad$ .

Use : $\log _{10} 2=0.3$

Answer: 15.62

Numerical answer — enter this value.

Step-by-step solution

\frac{X}{m} = K\, C^{1/n}

\log\!\left(\frac{X}{m}\right) = \log K + \frac{1}{n}\log C

\log 4 = \log K + \frac{1}{n}\log 10

0.6 = \log K + \frac{1}{n}

\log 10 = \log K + \frac{1}{n}\log 16

1 = \log K + \frac{1}{n}\times 1.2

Subtracting equation (1) from equation (2):

0.4 = \frac{1}{n}(0.2) \Rightarrow n = 0.5

\log K = -1.4

\log\!\left(\frac{X}{m}\right) = \log K + \frac{1}{n}\log C

= -1.4 + 2\log 20

= -1.4 + 2.6 = 1.2

\frac{X}{m} = 10^{1.2} = 16

(\log 2 = 0.3,\; 4\log 2 = 1.2,\; 16 = 10^{1.2})

\frac{X}{m} = K\, C^{1/n}

4 = K(10)^{1/n}

10 = K(16)^{1/n}

X = K(20)^{1/n}

Solving equations (1) and (2):

\frac{1}{n} = 2

Solving equations (1) and (3):

\frac{4}{X} = \left(\frac{10}{20}\right)^2 \Rightarrow X = 16

Solving equations (2) and (3):

\frac{10}{X} = \left(\frac{16}{20}\right)^2 \Rightarrow X = 15.625

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Chemistry
Chapter: Surface Chemistry
Topic: Adsorption