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 : $\log_{10} 2=0.3$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Surface Chemistry

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

10 mg g^{- 1}

and

16 mg g^{- 1}

aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be

4 mg g^{- 1}

and

10 mg g^{- 1}

, respectively. At this temperature, the concentration (in

mg g^{- 1}

) of adsorbed phenol from

20 mg g^{- 1}

aqueous solution of phenol will be ____.

Use :

lo g_{10} 2 = 0.3

Answer: 15.62

The Freundlich isotherm is:

\frac{x}{m} = K \times C^{1/ n}

Taking logarithms:

lo g (\frac{x}{m}) = lo g K + \frac{1}{n} lo g C

From the given data:

Equation 1:

lo g 4 = lo g K + \frac{1}{n} lo g 10

0.6 = lo g K + \frac{1}{n}

Equation 2:

lo g 10 = lo g K + \frac{1}{n} lo g 16

1 = lo g K + \frac{1}{n} \times 1.2

Subtracting equation 1 from equation 2:

0.4 = \frac{1}{n} \times 0.2

\frac{1}{n} = 2

Substituting back:

lo g K = 0.6 - 2 = - 1.4

For C = 20 mg/g:

lo g (\frac{x}{m}) = - 1.4 + 2 \times lo g 20

= - 1.4 + 2 (lo g 2 + lo g 10) = - 1.4 + 2 (0.3 + 1) = - 1.4 + 2.6 = 1.2

\frac{x}{m} = 1 0^{1.2} = 10 \times 1 0^{0.2} = 10 \times 2^{0.666...} \approx 16.00 mg/g

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer