JEE Advanced 2025 — Paper 2 — CHEM-III solution

JEE Advanced2025Paper 2CHEM-III

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 : $lo g_{10} 2 = 0.3$

Your answer

Solution

The Freundlich relation is $\frac{x}{m} = K C^{1/ n}$ . Taking the ratio of the two given data points eliminates $K$ :

$\frac{10}{4} = (\frac{16}{10})^{1/ n} \Rightarrow lo g 2.5 = \frac{1}{n} lo g 1.6$

Using $lo g 2 = 0.3$ : $lo g 2.5 = lo g (10/4) = 1 - 2 lo g 2 = 1 - 0.6 = 0.4$ and $lo g 1.6 = lo g (16/10) = 4 lo g 2 - 1 = 1.2 - 1 = 0.2$ . So $1/ n = 0.4/0.2 = 2$ .

Now from $\frac{x _{1}}{m} = K (C_{1})^{1/ n}$ : $4 = K (10)^{2}$ , giving $K = 4 \times 1 0^{- 2}$ .

For $C = 20$ mg g $^{- 1}$ :

$\frac{x}{m} = 4 \times 1 0^{- 2} \times (20)^{2} = 4 \times 1 0^{- 2} \times 400 = 16.00$ mg g $^{- 1}$ .

$∴ \frac{x}{m} = 16.00$ mg g $^{- 1}$ .

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