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Principles Related JEE 2024: Adsorption Answer Explained

JEE Advanced 2024 Chemistry Principles Related to Practical Chemistry Adsorption and monolayer surface area

By Founder, JEEnius - IIT Kanpur Alumni · Sep 29, 2026 · 4 min read

Medium 2 min target

Q.8 To form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P×10−23 m² surface area on charcoal, the value of P is _____.

Use given data: Surface area of charcoal =1.5×102 m² g⁻¹; Avogadro's number NA=6.0×1023 mol⁻¹.

Show answerAnswer

2500

Explanation

Initial moles of acetic acid used are calculated from molarity and volume.

Moles of acetic acid initially=M×V

Moles of acetic acid initially=0.5×100×10−3

Moles of acetic acid initially=5×10−2 mol

The unadsorbed acetic acid is neutralized by NaOH. Since acetic acid is monoprotic, it reacts with NaOH in a 1:1 mole ratio.

Moles of NaOH=1×40×10−3

Moles of NaOH=4×10−2 mol

Therefore, moles of unadsorbed acetic acid are:

Moles of unadsorbed acetic acid=4×10−2 mol

So, moles of acetic acid adsorbed on charcoal are:

Moles adsorbed=5×10−2−4×10−2

Moles adsorbed=1×10−2 mol

Number of acetic acid molecules adsorbed is:

N=1×10−2×6.0×1023

N=6.0×1021

Total surface area of 1 g charcoal is:

A=1.5×102 m²

Area occupied by one molecule is:

A1=1.5×1026.0×1021

A1=0.25×10−19 m²

A1=2.5×10−20 m²

Given that area per molecule is P×10−23 m², compare:

P×10−23=2.5×10−20

P=2.5×103

P=2500

Therefore, the value of P is 2500.

Chemistry artwork for the article: Principles Related JEE 2024: Adsorption Answer Explained

What is the answer to the Principles Related JEE 2024 adsorption question?

The numerical entry is 2500, not the area per molecule. The Principles Related JEE 2024 adsorption question is JEE Advanced 2024, Paper 2, Chemistry, Q.8, from Principles Related to Practical Chemistry. It is a numerical-answer question, not an MCQ.

The question bank rates it medium and gives an expected solving time of 120 seconds. These are question-bank tags, not official exam classifications.

A 1 g charcoal sample develops a complete acetic-acid monolayer after contact with 100 mL of 0.5 M acid. The acid remaining in solution needs 40 mL of 1 M NaOH for neutralisation.

The supplied specific surface area and Avogadro constant are:

s=1.5×102 m2g−1,NA=6.0×1023 mol−1

Find the numerical coefficient in the stated area occupied by each molecule:

A1=P×10−23 m2

How does the titration tell us how much acid did not adsorb?

The NaOH measures unadsorbed acid remaining in solution. Subtract that amount from the initial acid to obtain the amount on the charcoal. The calculation must keep these two acid inventories separate: acid in the liquid and acid on the surface.

First convert both volumes to litres:

100 mL=100×10−3 L
40 mL=40×10−3 L

Calculate the initial acid from molarity and volume:

ninitial=MV=0.5×100×10−3=5×10−2 mol

Acetic acid reacts with sodium hydroxide as follows:

CH3COOH+NaOH→CH3COONa+H2O

Acetic acid is monoprotic: each molecule supplies one acidic proton for neutralisation. The acid-to-base mole ratio is therefore one to one.

nNaOH=1×40×10−3=4×10−2 mol

Hence the acid remaining in the liquid is:

nunadsorbed=nNaOH=4×10−2 mol

Now subtract to obtain the adsorbed acid:

nadsorbed&=ninitial−nunadsorbed&=5×10−2−4×10−2&=1×10−2 mol

Weak-acid behaviour does not change the neutralisation mole ratio. It concerns ionisation in water, not the amount of NaOH needed for complete neutralisation. No degree-of-ionisation correction belongs in this calculation.

How do adsorbed moles give the area occupied by one molecule?

Divide the total charcoal surface area by the number of adsorbed molecules. Complete monolayer coverage means those molecules collectively cover the stated surface in one layer. Only the adsorbed amount belongs in the molecule count.

Following the official sequence, first convert adsorbed moles to molecules:

N=nadsorbedNA=1×10−2×6.0×1023=6.0×1021 molecules

Then multiply specific surface area by charcoal mass:

A=(1.5×102 m2g−1)(1 g)=1.5×102 m2

Keep the three area quantities separate:

  • Specific surface area: surface area per gram of charcoal.
  • Total surface area: surface area of the full charcoal sample.
  • Area per molecule: covered area divided by the adsorbed molecule count.

For a complete monolayer, the adsorbed molecules cover the entire stated surface: A=NA1

Therefore:

A1=AN=1.5×1026.0×1021=0.25×10−19=2.5×10−20 m2

The result is an area per molecule, not an area per gram. No molecular shape or adsorption-isotherm model is needed.

Why is the numerical entry 2500 rather than the molecular area?

The requested answer is a dimensionless coefficient multiplying the given area scale. Match the calculated molecular area to the form specified in the question:

P×10−23 m2=2.5×10−20 m2
P=2.5×10−2010−23=2.5×103=2500

Enter:

2500

The square-metre units cancel when finding the coefficient. By contrast, the molecular area retains area units:

A1=2.5×10−20 m2

Reverse-check the exponent conversion:

2500×10−23 m2=2.5×10−20 m2

Multiplying this area per molecule by the adsorbed molecule count must return the total charcoal area:

(6.0×1021)(2.5×10−20 m2)=150 m2

Why would using the acid left in solution give 625?

That substitution counts molecules in the liquid as the monolayer. There are no wrong options in the supplied numerical-answer question. The value 625 below is a derived incorrect result, not an official distractor.

Using the unadsorbed amount gives:

Nwrong=0.04×6.0×1023=2.4×1022 molecules

Continuing that error:

A1,wrong=1502.4×1022=6.25×10−21 m2
Pwrong=6.25×10−2110−23=625

The arithmetic is consistent, but the denominator counts molecules in the liquid, not molecules on the charcoal. Check the ratio of the wrongly used amount to the actual adsorbed amount:

0.040.01=4

Four times too many molecules makes the calculated area per molecule one-quarter of the correct value. The same factor carries into the numerical coefficient.

Before substituting numbers, label the initial, unadsorbed and adsorbed amounts separately. Use these labels throughout the calculation:

ninitial,nunadsorbed,nadsorbed

What three follow-up questions test this method?

Use these exercises to test the mole balance, neutralisation and coverage assumption. They are original follow-up practice based on this PYQ, not questions from another JEE paper. Each uses the supplied data, with its worked answer immediately below.

1. What percentage of the initial acetic acid was adsorbed?

20% of the initial acid was adsorbed. Divide the adsorbed amount by the initial amount:

Percentage adsorbed=0.010.05×100=20%

Only 20% of the initial acid adsorbs, yet the charcoal has complete surface coverage. Percentage adsorption measures the fraction of acid removed from solution, not the fraction of charcoal surface covered.

2. What volume of the same 1 M NaOH would neutralise the recovered adsorbed acid?

10 mL is required, assuming complete recovery with the acid unchanged. The one-to-one neutralisation ratio gives:

nNaOH=0.01 mol
V=nM=0.011=0.01 L=10 mL

3. Would dividing total area by molecule count work for incomplete single-layer coverage?

No: including uncovered charcoal surface would overestimate the occupied area per molecule. The correct numerator would be only the covered area:

A1,true=AcoveredN

Before dividing area by molecule count, underline complete monolayer in the question. That condition justifies using the entire charcoal surface.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Related on JEEnius: Motion in a Plane JEE 2022: Gravity-Change Solution.

Frequently asked questions

What is the answer to the Principles Related JEE 2024 adsorption question?

The numerical entry for JEE Advanced 2024 Paper 2 Chemistry Q.8 is 2500. The calculated area per molecule is 2.5 × 10^-20 m², but the question asks for P in A₁ = P × 10^-23 m². Dividing the molecular area by the stated area scale gives P = 2500.

How do I calculate the moles of acetic acid adsorbed on charcoal?

The initial acid amount is 0.5 × 0.100 = 0.05 mol. The remaining acid requires 0.040 mol of NaOH, so the one-to-one neutralisation ratio gives 0.040 mol of unadsorbed acid. Subtracting gives 0.05 − 0.04 = 0.01 mol adsorbed on charcoal.

Why am I getting 625 instead of 2500 in the adsorption question?

The result 625 comes from using the 0.04 mol of acid left in solution instead of the 0.01 mol adsorbed on charcoal. This counts four times too many molecules, making the calculated area per molecule and numerical coefficient one-quarter of their correct values. Only adsorbed molecules belong in the monolayer calculation.

Does acetic acid being a weak acid change the NaOH calculation?

No. Acetic acid is monoprotic and reacts with NaOH in a one-to-one mole ratio during complete neutralisation. Its weak-acid behaviour concerns ionisation in water, so no degree-of-ionisation correction is needed here.

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