Principles Related JEE 2025: Phenol Adsorption Solution

What is the phenol adsorption question in Principles Related JEE 2025?

The official answer to the Principles Related JEE 2025 phenol adsorption question is 16. It comes from JEE Advanced 2025, Paper 2, Chemistry, under Principles Related to Practical Chemistry. The question bank labels this numerical-answer problem medium and gives an expected solving time of 120 seconds. These are the question bank’s ratings, not an official difficulty classification or an exam-imposed time limit.

Phenol dissolved in water is adsorbed onto fly ash. At a fixed temperature, it follows the Freundlich isotherm, with these measured concentrations:

First measurement:

C = 10 m g g^{- 1}, \frac{x}{m} = 4 m g g^{- 1}

Second measurement:

C = 16 m g g^{- 1}, \frac{x}{m} = 10 m g g^{- 1}

Find the adsorbed phenol concentration, in milligrams per gram, at the following aqueous concentration. Use the supplied logarithm approximation.

C = 20 m g g^{- 1}

\log_{10} 2 = 0.3

How do you write the Freundlich isotherm and eliminate the constant?

Divide the second measurement equation by the first. This removes the Freundlich constant before you evaluate any logarithms, leaving one unknown exponent. Finding the constant first would only add algebra.

The Freundlich adsorption isotherm is:

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

The left side is the adsorbed phenol concentration on fly ash. The concentration raised to a power on the right is the equilibrium aqueous phenol concentration. Both retain the question’s units of milligrams per gram: $m g g^{- 1}$

For this adsorption system at the stated fixed temperature, both parameters remain unchanged:

k, n

Substitute the first and second measurements: $4 = k \times 10^{1 / n}$ $10 = k \times 16^{1 / n}$

Divide with the second measurement in the numerator on both sides. The constant cancels:

\frac{10}{4} = \frac{k \times 16^{1 / n}}{k \times 10^{1 / n}} = {(\frac{16}{10})}^{1 / n}

2.5 = {1.6}^{1 / n}

How do you calculate the exponent using the supplied logarithm?

The exponent is 2 under the supplied logarithm approximation. Keep that approximation throughout the official calculation. Calculator-precision logarithms give a different exponent.

Take base-10 logarithms of the ratio equation. The power rule brings the exponent outside:

\log_{10} 2.5 = \frac{1}{n} \log_{10} 1.6

Use the quotient rule to obtain the logarithm of 5. The base-10 logarithm of 10 is 1:

\log_{10} 5 = \log_{10} (\frac{10}{2}) = 1 - \log_{10} 2 = 1 - 0.3 = 0.7

Apply the same quotient rule to 2.5:

\log_{10} 2.5 = \log_{10} (\frac{5}{2}) = 0.7 - 0.3 = 0.4

For 1.6, express the numerator as a power of 2. Then apply the quotient and power rules:

\log_{10} 1.6 = \log_{10} (\frac{2^{4}}{10}) = 4 \log_{10} 2 - 1 = 4 \times 0.3 - 1 = 0.2

Substitute these approximate logarithms:

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

\frac{1}{n} = \frac{0.4}{0.2} = 2

Do not confuse the exponent with the parameter itself. The power applied to concentration is 2, not one-half:

\frac{1}{n} = 2, n = \frac{1}{2}

How does the official solution obtain the answer 16?

Return to the first measurement, as the official solution does. With the exponent fixed at 2, this gives the Freundlich constant as 0.04 in the concentration convention used by the question: $4 = k \times 10^{2}$

k = \frac{4}{100} = 0.04

Substitute the requested aqueous concentration:

\frac{x}{m} = 0.04 \times 20^{2} = 0.04 \times 400 = 16 m g g^{- 1}

The physical result above is in milligrams per gram. The official numerical entry is:

16

Within this adopted calculation, doubling the aqueous concentration multiplies adsorption by the square of 2. Thus the original adsorbed concentration of 4 becomes 16:

\frac{20}{10} = 2, 2^{2} = 4

This checks the substitution, not the precision of the rounded logarithms.

Why does another calculation give 15.625?

The result 15.625 comes from using the second measurement after adopting the approximate exponent of 2. This is a numerical-answer question with no listed options: 15.625 is a calculated result that differs from the official answer, not a supplied distractor.

Using the second measurement to determine the constant gives:

k = \frac{10}{16^{2}} = \frac{10}{256} = 0.0390625

\frac{x}{m} = 0.0390625 \times 20^{2} = 0.0390625 \times 400 = 15.625 m g g^{- 1}

The multiplication is correct. This route changes the calibration point after adopting an exponent obtained from the supplied approximate logarithm. It does not reproduce the official solution’s first-measurement calibration.

The official constant and exponent do not fit the second measurement exactly. At that aqueous concentration, they give:

0.04 \times 16^{2} = 10.24 m g g^{- 1}

That is not exactly the stated adsorbed concentration of 10. The approximate exponent cannot fit both supplied measurements with one constant.

For two exact measurements, the exactly fitted exponent uses unrounded logarithms:

\frac{1}{n} = \frac{\log_{10} (2.5)}{\log_{10} (1.6)}

With that exponent, either measurement determines the same constant. Using the second measurement is not inherently invalid mathematics. Here, 16 is the official answer from the stipulated worked route; the discrepancy reflects sensitivity to the supplied approximation.

Which two practice questions check the Freundlich method?

Check the slope of the logarithmic plot and calculate adsorption directly from a concentration ratio. Both items below are original practice based on this question, not additional verified PYQs.

Question 1: What is the slope of the logarithmic adsorption plot?

The slope is 2 under the official calculation. Plot the logarithm of adsorbed concentration vertically against the logarithm of aqueous concentration:

\log_{10} (\frac{x}{m}) = \log_{10} k + \frac{1}{n} \log_{10} C

The coefficient of the horizontal-axis quantity gives:

slope = \frac{1}{n} = 2

Question 2: How can you calculate adsorption at 20 from the measurement at 10 without finding the constant again?

Use the first measurement as a reference and take a concentration ratio. This shortcut requires the exponent already obtained in the official calculation:

\frac{{(\frac{x}{m})}_{C = 20}}{4} = {(\frac{20}{10})}^{2} = 4

{(\frac{x}{m})}_{C = 20} = 4 \times 4 = 16 m g g^{- 1}

On your next attempt, write “rounded exponent, first measurement” beside the substitution. That keeps the approximation and calibration choice explicit.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Keep going with Oscillations and Waves JEE 2021: Answer Verification.

Frequently asked questions

What is the answer to the JEE Advanced 2025 phenol adsorption question?

The official numerical answer is 16, corresponding to an adsorbed phenol concentration of 16 mg/g. Using the supplied logarithm approximation gives the Freundlich exponent 1/n = 2. Substituting the first measurement gives k = 0.04, so at C = 20, x/m = 0.04 × 20² = 16 mg/g.

How do I eliminate k in the Freundlich adsorption equation?

Write x/m = kC^(1/n) for each measurement and divide the second equation by the first. For this question, 10/4 = (16/10)^(1/n), so k cancels and only the exponent remains unknown.

Is n or 1/n equal to 2 in the phenol adsorption question?

The exponent 1/n equals 2 under the supplied approximation log10(2) = 0.3; n equals 1/2. The calculation gives 1/n = log10(2.5)/log10(1.6) = 0.4/0.2 = 2. This exponent is also the slope of the plot of log10(x/m) against log10(C).

Why do I get 15.625 instead of 16 in the phenol adsorption question?

Using the second measurement with the approximate exponent 2 gives k = 10/16² and predicts 15.625 mg/g at C = 20. The official solution instead uses the first measurement, giving k = 0.04 and an answer of 16 mg/g. The approximate exponent cannot fit both measurements exactly, so the discrepancy reflects approximation sensitivity rather than invalid mathematics.