Use Freundlich adsorption isotherm

The Freundlich isotherm is $\frac{x}{m} = k C^{1/n}$ where:

• $\frac{x}{m}$ = amount adsorbed per gram of adsorbent
• $C$ = concentration of solution
• $k, n$ = constants

Given data

For $C_1 = 10\ \mathrm{mg\ g^{-1}}$, $\left(\frac{x}{m}\right)_1 = 4\ \mathrm{mg\ g^{-1}}$

For $C_2 = 16\ \mathrm{mg\ g^{-1}}$, $\left(\frac{x}{m}\right)_2 = 10\ \mathrm{mg\ g^{-1}}$

Form ratio to eliminate $k$

$\frac{10}{4} = \left(\frac{16}{10}\right)^{1/n}$

$2.5 = (1.6)^{1/n}$

Taking base-10 logarithm: $\log 2.5 = \frac{1}{n} \log 1.6$

Evaluate logarithms using $\log 2 = 0.3$

$\log 2.5 = \log \left(\frac{5}{2}\right) = \log 5 - \log 2$

Since $\log 5 = \log 10 - \log 2 = 1 - 0.3 = 0.7$ so $\log 2.5 = 0.7 - 0.3 = 0.4$

Also, $\log 1.6 = \log \left(\frac{16}{10}\right) = \log 16 - 1$ and $\log 16 = \log(2^4) = 4\log 2 = 4(0.3)=1.2$ hence $\log 1.6 = 1.2 - 1 = 0.2$

Therefore, $0.4 = \frac{1}{n}(0.2)$ $\frac{1}{n} = 2$

Find $k$

Using $\frac{x}{m} = k C^2$

from the first data point: $4 = k(10)^2$ $k = \frac{4}{100} = 0.04$

Find adsorption at $C = 20\ \mathrm{mg\ g^{-1}}$

$\frac{x}{m} = 0.04(20)^2 = 0.04 \times 400 = 16$

Final answer

The concentration of adsorbed phenol is $\boxed{16\ \mathrm{mg\ g^{-1}}}$