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 : \(\log _{10} 2=0.3\)

\(\begin{aligned}
& \frac{x}{m}=K \times C^{1 / n} \\
& \log \left(\frac{x}{m}\right)=\log K+\frac{1}{n} \log C \\
& \log 4=\log K+\frac{1}{n} \log 10 \\
& 0.6=\log K+\frac{1}{n} .....(1) \\
& \log 10=\log K+\frac{1}{n} \log 16 \\
& 1=\log K+\frac{1}{n} \times 1.2 ....(2) \end{aligned}\)
Equation (2) - equation (1)
\(0.4=\frac{1}{\mathrm{n}} \times(0.2) \Rightarrow \mathrm{n}=0.5\)
and \(\log \mathrm{K}=-1.4\)
\(\begin{aligned}
& \log \frac{x}{m}=\log K+\frac{1}{n} \times \log C \\
& =-1.4+2 \times \log 20
\end{aligned}\)
\(\begin{aligned} & =-1.4+2.6=1.2 \\ & \frac{\mathrm{x}}{\mathrm{m}}=10^{+1.2}=16 \\ & \left(\log 2=0.3,4 \log 2=1.2,16=10^{+1.2}\right)\end{aligned}\)
\(\begin{aligned}
& \frac{x}{m}=K \times C^{1 / n} \\
& 4=K(10)^{1 / n}....(1) \\
& 10=K(16)^{1 / n}....(2) \\
& X=K(20)^{1 / n}....(3)
\end{aligned}\)
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)
\(\begin{aligned}
& \frac{10}{X}=\left(\frac{16}{20}\right)^2 \\
& X=15.625
\end{aligned}\)