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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Surface Chemistry

JEE Advanced 2025 · Paper 2 · Chemistry · Surface Chemistry

Adsorption of phenol from its aqueous solution on to fly ash obeys...

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10mgg−110 \mathrm{mg} \mathrm{g}^{-1} and 16mgg−116 \mathrm{mg} \mathrm{g}^{-1} aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4mgg−14 \mathrm{mg} \mathrm{g}^{-1} and 10mgg−110 \mathrm{mg} \mathrm{g}^{-1}, respectively. At this temperature, the concentration (in mgg−1\mathrm{mg} \mathrm{g}^{-1} ) of adsorbed phenol from 20mgg−120 \mathrm{mg} \mathrm{g}^{-1} aqueous solution of phenol will be \qquad .
Use : log⁡102=0.3\log _{10} 2=0.3

Correct answer

15.62

Step-by-step solution

xm=K×C1/nlog⁡(xm)=log⁡K+1nlog⁡Clog⁡4=log⁡K+1nlog⁡100.6=log⁡K+1n.....(1)log⁡10=log⁡K+1nlog⁡161=log⁡K+1n×1.2....(2)\begin{aligned} & \frac{x}{m}=K \times 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} .....(1) \\ & \log 10=\log K⁺\frac{1}{n} \log 16 \\ & 1=\log K⁺\frac{1}{n} \times 1.2 ....(2) \end{aligned} Equation (2) - equation (1) 0.4=1n×(0.2)⇒n=0.50.4=\frac{1}{\mathrm{n}} \times(0.2) \Rightarrow \mathrm{n}=0.5 and log⁡K=−1.4\log \mathrm{K}=-1.4log⁡xm=log⁡K+1n×log⁡C=−1.4+2×log⁡20\begin{aligned} & \log \frac{x}{m}=\log K⁺\frac{1}{n} \times \log C \\ & =-1.4+2 \times \log 20 \end{aligned}=−1.4+2.6=1.2xm=10+1.2=16(log⁡2=0.3,4log⁡2=1.2,16=10+1.2)\begin{aligned} & =-1.4+2.6=1.2 \\ & \frac{\mathrm{x}}{\mathrm{m}}=10^{+1.2}=16 \\ & \left(\log 2=0.3,4 \log 2=1.2,16=10^{+1.2}\right)\end{aligned}xm=K×C1/n4=K(10)1/n....(1)10=K(16)1/n....(2)X=K(20)1/n....(3)\begin{aligned} & \frac{x}{m}=K \times C^{1 / n} \\ & 4=K(10)^{1 / n}....(1) \\ & 10=K(16)^{1 / n}....(2) \\ & X=K(20)^{1 / n}....(3) \end{aligned}On solving equation (1) and (2) $\frac{1}{n}=2$ On solving equation (1) and (3) $\frac{4}{X}=\left(\frac{10}{20}\right)^2 X=16$ On solving equation (2) and (3)\begin{aligned} & \frac{10}{X}=\left(\frac{16}{20}\right)^2 \\ & X=15.625 \end{aligned}$$
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