JEE Advance - Chemistry (2025 - Paper 2 Online - No. 11)
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 ______________.
Use: $\log _{10} 2=0.3$
Explanation
$\begin{aligned} & \frac{x}{m}=K(C)^{\frac{1}{n}} \\\\ & 4=K(10)^{\frac{1}{n}}........(i) \\\\ & 10=K(16)^{\frac{1}{n}}.......(ii)\end{aligned}$
From (ii) and (i)
$$ \begin{aligned} & \log 4-\log 10=\frac{1}{n}(\log 10-\log 16) \\ & 2 \log 2-1=\frac{1}{n}(1-4 \log 2) \\ & 0.6-1=\frac{1}{x}(1-1.2) \\ & -0.4=\frac{-0.2}{n} \\ & n=\frac{2}{4}=\left(\frac{1}{2}\right) \end{aligned} $$
$\begin{aligned} \text { Now } 10 & =K(16)^2 \\ K & =\frac{10}{256} \\ \text { So, } \frac{x}{m} & =K(20)^2 \\ \frac{x}{m} & =\frac{10}{256} \times 400 \\ \frac{x}{m} & =15.625\end{aligned}$
or
$$ \begin{aligned} & 4=K(10)^2 \\ & K=\frac{4}{100} \\ & \frac{x}{m}=K(20)^2 \\ & \frac{x}{m}=\frac{4}{100} \times 400 \\ & \frac{x}{m}=16 \end{aligned} $$
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