$\text{Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm.}\text{At a given temperature,}$

$\text{from}\mathrm{ 10}\mathrm{mg}{\mathrm{ g}}^{-1}\text{and}\mathrm{ 16}\mathrm{mg}{g}^{-1}\text{aqueous phenol solutions, the concentrations of adsorbed phenol are measured to }$

$\text{be }4\mathrm{ mg}{\mathrm{ g}}^{-1}\text{and}\mathrm{ 10}\mathrm{ m\; g}{\mathrm{ g}}^{-1},\text{respectively.}\text{ At this temperature, the concentration (in}\mathrm{mg}{g}^{-1}\text{) of adsorbed}$

$\text{phenol from }20\mathrm{ m\; g}{\mathrm{ g}}^{-1}\text{aqueous solution of phenol will be}\text{\_\_\_\_\_\_}$
$\text{Use: }{\log }_{10}2=0.3$

The correct answer is 16.00.

Step-by-Step Explanation:

According to Freundlich adsorption isotherm:
$\frac{x}{m}=k{C}^{1/n}$
where:
• $\frac{x}{m}$ is the concentration of adsorbed phenol (adsorbed amount per unit mass of fly ash in mg g^-1).
• $C$ is the concentration of phenol in the aqueous solution (in mg g^-1 or mg L^-1 as given in the problem context).
• $k$ and $n$ are empirical constants.

Taking the logarithm on both sides:
$\log \left(\frac{x}{m}\right)=\log k+\frac{1}{n}\log C$

Let us denote ${\left(\frac{x}{m}\right)}_1=4$ mg g^-1 for solution concentration $C_1=10$ mg g^-1.
And ${\left(\frac{x}{m}\right)}_2=10$ mg g^-1 for solution concentration $C_2=16$ mg g^-1.

Formulating the two equations:
Equation (1):
$\log \left(4\right)=\log k+\frac{1}{n}\log \left(10\right)$
Equation (2):
$\log \left(10\right)=\log k+\frac{1}{n}\log \left(16\right)$

Subtracting Equation (1) from Equation (2):
$\log \left(10\right)-\log \left(4\right)=\frac{1}{n}\left[\log \left(16\right)-\log \left(10\right)\right]$

Using logarithmic identities:
• $\log \left(10\right)=1$
• $\log \left(4\right)=2\log \left(2\right)=2\times 0.3=0.6$
• $\log \left(16\right)=4\log \left(2\right)=4\times 0.3=1.2$

Substitute these values into the subtraction step:
$1-0.6=\frac{1}{n}\left[1.2-1\right]$
$0.4=\frac{1}{n}\left[0.2\right]$
$\frac{1}{n}=\frac{0.4}{0.2}=2$

Now, find the value of $\log k$ using Equation (1):
$\log \left(4\right)=\log k+2\log \left(10\right)$
$0.6=\log k+2\left(1\right)$
$\log k=0.6-2=-1.4$

Let ${\left(\frac{x}{m}\right)}_3$ be the adsorbed amount of phenol at solution concentration $C_3=20$ mg g^-1:
$\log {\left(\frac{x}{m}\right)}_3=\log k+\frac{1}{n}\log \left(20\right)$
Since $\log \left(20\right)=\log (10\times 2)=\log \left(10\right)+\log \left(2\right)=1+0.3=1.3$:

$\log {\left(\frac{x}{m}\right)}_3=-1.4+2.6=1.2$

Since $1.2=4\times 0.3=4\log \left(2\right)=\log \left(2^4\right)=\log \left(16\right)$:
${\left(\frac{x}{m}\right)}_3=16$

Thus, the concentration of adsorbed phenol from a 20 mg g^-1 aqueous solution is 16.00 mg g^-1.