JEE Advanced Previous Year Questions (2021 - 2024): Surface Chemistry | Chemistry for JEE Main & Advanced PDF Download

 Page 1


 
Q1:  T o form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5M acetic acid was used.
Some of the acetic acid remained unadsorbed. T o neutralize the unadsorbed acetic acid, 40 mL of 1M NaOH
solution was required. If each molecule of acetic acid occupies P × 10 ?²³ m² surface area on charcoal, the
value of P is ______.      [JEE Advanced 2024 Paper 2]
[Use given data: Surface area of charcoal = 1.5 × 10² m² g ?¹; Avogadro's number (N ?) = 6.0 × 10²³ mol ?¹] 
Ans: 2500
Let's rst understand and break down the given information. We are provided with the following details to solve
for the value of P:
1. Volume of acetic acid solution used: 100 mL of 0.5M acetic acid.
2. Volume of NaOH solution used for neutralization: 40 mL of 1M NaOH.
3. Surface area of the charcoal: 1.5 × 10² m² g ?¹.
4. Avogadro's number: N = 6.0 × 10²³ mol ?¹.
Step 1: Calculate the Moles of Acetic Acid Used
Moles of acetic acid initially = 0.5 M × 
100 mL
1000
 = 0.05 moles
Step 2: Calculate the Moles of NaOH Used for Neutralization
Moles of NaOH = 1 M × 
40 mL
1000
 = 0.04 moles
Since NaOH completely neutralizes the unadsorbed acetic acid:
Moles of unadsorbed acetic acid = 0.04 moles
Step 3: Calculate the Moles of Adsorbed Acetic Acid
Moles of adsorbed acetic acid = 0.05 - 0.04 = 0.01 moles
Step 4: Calculate the T otal Number of Acetic Acid Molecules Adsorbed
Number of molecules = 0.01 moles × 6.0 × 10²³ molecules/mole = 6.0 × 10²¹ molecules
Step 5: Calculate the T otal Surface Area
T otal surface area = 1.5 × 10² m²/g × 1 g = 1.5 × 10² m²
Step 6: Determine the Area Occupied by Each Molecule of Acetic Acid
Area per molecule of acetic acid = 
T otal surface area
Number of molecules
 = 
1.5 × 10² m²
6.0 × 10²¹ molecules
Step 7: Perform the Calculation
Area per molecule of acetic acid = 
1.5 × 10²
6.0 × 10²¹
 = 0.25 × 10 ?¹ ? m² = 2.5 × 10 ?² ° m²
Since the problem states that the area per molecule is P × 10 ?²³ m², we equate:
2.5 × 10 ?²° = P × 10 ?²³
A
Page 2


 
Q1:  T o form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5M acetic acid was used.
Some of the acetic acid remained unadsorbed. T o neutralize the unadsorbed acetic acid, 40 mL of 1M NaOH
solution was required. If each molecule of acetic acid occupies P × 10 ?²³ m² surface area on charcoal, the
value of P is ______.      [JEE Advanced 2024 Paper 2]
[Use given data: Surface area of charcoal = 1.5 × 10² m² g ?¹; Avogadro's number (N ?) = 6.0 × 10²³ mol ?¹] 
Ans: 2500
Let's rst understand and break down the given information. We are provided with the following details to solve
for the value of P:
1. Volume of acetic acid solution used: 100 mL of 0.5M acetic acid.
2. Volume of NaOH solution used for neutralization: 40 mL of 1M NaOH.
3. Surface area of the charcoal: 1.5 × 10² m² g ?¹.
4. Avogadro's number: N = 6.0 × 10²³ mol ?¹.
Step 1: Calculate the Moles of Acetic Acid Used
Moles of acetic acid initially = 0.5 M × 
100 mL
1000
 = 0.05 moles
Step 2: Calculate the Moles of NaOH Used for Neutralization
Moles of NaOH = 1 M × 
40 mL
1000
 = 0.04 moles
Since NaOH completely neutralizes the unadsorbed acetic acid:
Moles of unadsorbed acetic acid = 0.04 moles
Step 3: Calculate the Moles of Adsorbed Acetic Acid
Moles of adsorbed acetic acid = 0.05 - 0.04 = 0.01 moles
Step 4: Calculate the T otal Number of Acetic Acid Molecules Adsorbed
Number of molecules = 0.01 moles × 6.0 × 10²³ molecules/mole = 6.0 × 10²¹ molecules
Step 5: Calculate the T otal Surface Area
T otal surface area = 1.5 × 10² m²/g × 1 g = 1.5 × 10² m²
Step 6: Determine the Area Occupied by Each Molecule of Acetic Acid
Area per molecule of acetic acid = 
T otal surface area
Number of molecules
 = 
1.5 × 10² m²
6.0 × 10²¹ molecules
Step 7: Perform the Calculation
Area per molecule of acetic acid = 
1.5 × 10²
6.0 × 10²¹
 = 0.25 × 10 ?¹ ? m² = 2.5 × 10 ?² ° m²
Since the problem states that the area per molecule is P × 10 ?²³ m², we equate:
2.5 × 10 ?²° = P × 10 ?²³
A
Step 8: Solve for P
P = 
2.5 × 10 ?² °
10 ?²³
 = 2.5 × 10³ = 2500
Therefore, the value of P is 2500.
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Q1:  T o form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5M acetic acid was used.
Some of the acetic acid remained unadsorbed. T o neutralize the unadsorbed acetic acid, 40 mL of 1M NaOH
solution was required. If each molecule of acetic acid occupies P × 10 ?²³ m² surface area on charcoal, the
value of P is ______.      [JEE Advanced 2024 Paper 2]
[Use given data: Surface area of charcoal = 1.5 × 10² m² g ?¹; Avogadro's number (N ?) = 6.0 × 10²³ mol ?¹] 
Ans: 2500
Let's rst understand and break down the given information. We are provided with the following details to solve
for the value of P:
1. Volume of acetic acid solution used: 100 mL of 0.5M acetic acid.
2. Volume of NaOH solution used for neutralization: 40 mL of 1M NaOH.
3. Surface area of the charcoal: 1.5 × 10² m² g ?¹.
4. Avogadro's number: N = 6.0 × 10²³ mol ?¹.
Step 1: Calculate the Moles of Acetic Acid Used
Moles of acetic acid initially = 0.5 M × 
100 mL
1000
 = 0.05 moles
Step 2: Calculate the Moles of NaOH Used for Neutralization
Moles of NaOH = 1 M × 
40 mL
1000
 = 0.04 moles
Since NaOH completely neutralizes the unadsorbed acetic acid:
Moles of unadsorbed acetic acid = 0.04 moles
Step 3: Calculate the Moles of Adsorbed Acetic Acid
Moles of adsorbed acetic acid = 0.05 - 0.04 = 0.01 moles
Step 4: Calculate the T otal Number of Acetic Acid Molecules Adsorbed
Number of molecules = 0.01 moles × 6.0 × 10²³ molecules/mole = 6.0 × 10²¹ molecules
Step 5: Calculate the T otal Surface Area
T otal surface area = 1.5 × 10² m²/g × 1 g = 1.5 × 10² m²
Step 6: Determine the Area Occupied by Each Molecule of Acetic Acid
Area per molecule of acetic acid = 
T otal surface area
Number of molecules
 = 
1.5 × 10² m²
6.0 × 10²¹ molecules
Step 7: Perform the Calculation
Area per molecule of acetic acid = 
1.5 × 10²
6.0 × 10²¹
 = 0.25 × 10 ?¹ ? m² = 2.5 × 10 ?² ° m²
Since the problem states that the area per molecule is P × 10 ?²³ m², we equate:
2.5 × 10 ?²° = P × 10 ?²³
A
Step 8: Solve for P
P = 
2.5 × 10 ?² °
10 ?²³
 = 2.5 × 10³ = 2500
Therefore, the value of P is 2500.
 
 
Single Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q1. Consider the following statements related to colloids. 
(I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion 
medium. 
(II) For emulsions, both the dispersed phase and the dispersion medium are liquid. 
(III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. 
(IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the 
same refractive index as that of the dispersion medium. 
The option with the correct set of statements is : 
A. (I) and (II) 
B. (II) and (III) 
C. (III) and (IV) 
D. (II) and (IV) 
Ans: (A) 
The correct answer is (A) (I) and (II). Here's why : 
(I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion medium. This 
statement is correct. Lyophobic colloids are "solvent-hating," and do not form easily. They require 
specific conditions and often require the use of special techniques like dispersion or condensation 
methods. 
(II) For emulsions, both the dispersed phase and the dispersion medium are liquid. This statement is 
also correct. An emulsion is a type of colloid where both phases are liquids. 
(III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. This statement 
is incorrect. Micelles only form when the concentration of the surfactant is above a certain threshold 
called the critical micelle concentration (CMC), and typically, this occurs at certain temperatures as 
well. 
(IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the same 
refractive index as that of the dispersion medium. This statement is incorrect. The Tyndall effect, which 
is the scattering of light by particles in a colloid or in a very fine suspension, is only observed when the 
particles have a different refractive index than the surrounding medium. If the refractive index is the 
same, the light will pass through without scattering and the Tyndall effect will not be observed. 
So, options (I) and (II) are correct and hence, the answer is (A) (I) and (II). 
 
 
 
Page 4


 
Q1:  T o form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5M acetic acid was used.
Some of the acetic acid remained unadsorbed. T o neutralize the unadsorbed acetic acid, 40 mL of 1M NaOH
solution was required. If each molecule of acetic acid occupies P × 10 ?²³ m² surface area on charcoal, the
value of P is ______.      [JEE Advanced 2024 Paper 2]
[Use given data: Surface area of charcoal = 1.5 × 10² m² g ?¹; Avogadro's number (N ?) = 6.0 × 10²³ mol ?¹] 
Ans: 2500
Let's rst understand and break down the given information. We are provided with the following details to solve
for the value of P:
1. Volume of acetic acid solution used: 100 mL of 0.5M acetic acid.
2. Volume of NaOH solution used for neutralization: 40 mL of 1M NaOH.
3. Surface area of the charcoal: 1.5 × 10² m² g ?¹.
4. Avogadro's number: N = 6.0 × 10²³ mol ?¹.
Step 1: Calculate the Moles of Acetic Acid Used
Moles of acetic acid initially = 0.5 M × 
100 mL
1000
 = 0.05 moles
Step 2: Calculate the Moles of NaOH Used for Neutralization
Moles of NaOH = 1 M × 
40 mL
1000
 = 0.04 moles
Since NaOH completely neutralizes the unadsorbed acetic acid:
Moles of unadsorbed acetic acid = 0.04 moles
Step 3: Calculate the Moles of Adsorbed Acetic Acid
Moles of adsorbed acetic acid = 0.05 - 0.04 = 0.01 moles
Step 4: Calculate the T otal Number of Acetic Acid Molecules Adsorbed
Number of molecules = 0.01 moles × 6.0 × 10²³ molecules/mole = 6.0 × 10²¹ molecules
Step 5: Calculate the T otal Surface Area
T otal surface area = 1.5 × 10² m²/g × 1 g = 1.5 × 10² m²
Step 6: Determine the Area Occupied by Each Molecule of Acetic Acid
Area per molecule of acetic acid = 
T otal surface area
Number of molecules
 = 
1.5 × 10² m²
6.0 × 10²¹ molecules
Step 7: Perform the Calculation
Area per molecule of acetic acid = 
1.5 × 10²
6.0 × 10²¹
 = 0.25 × 10 ?¹ ? m² = 2.5 × 10 ?² ° m²
Since the problem states that the area per molecule is P × 10 ?²³ m², we equate:
2.5 × 10 ?²° = P × 10 ?²³
A
Step 8: Solve for P
P = 
2.5 × 10 ?² °
10 ?²³
 = 2.5 × 10³ = 2500
Therefore, the value of P is 2500.
 
 
Single Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q1. Consider the following statements related to colloids. 
(I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion 
medium. 
(II) For emulsions, both the dispersed phase and the dispersion medium are liquid. 
(III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. 
(IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the 
same refractive index as that of the dispersion medium. 
The option with the correct set of statements is : 
A. (I) and (II) 
B. (II) and (III) 
C. (III) and (IV) 
D. (II) and (IV) 
Ans: (A) 
The correct answer is (A) (I) and (II). Here's why : 
(I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion medium. This 
statement is correct. Lyophobic colloids are "solvent-hating," and do not form easily. They require 
specific conditions and often require the use of special techniques like dispersion or condensation 
methods. 
(II) For emulsions, both the dispersed phase and the dispersion medium are liquid. This statement is 
also correct. An emulsion is a type of colloid where both phases are liquids. 
(III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. This statement 
is incorrect. Micelles only form when the concentration of the surfactant is above a certain threshold 
called the critical micelle concentration (CMC), and typically, this occurs at certain temperatures as 
well. 
(IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the same 
refractive index as that of the dispersion medium. This statement is incorrect. The Tyndall effect, which 
is the scattering of light by particles in a colloid or in a very fine suspension, is only observed when the 
particles have a different refractive index than the surrounding medium. If the refractive index is the 
same, the light will pass through without scattering and the Tyndall effect will not be observed. 
So, options (I) and (II) are correct and hence, the answer is (A) (I) and (II). 
 
 
 
2022 
Multiple Correct Type 
JEE Advanced 2022 Paper 1 Online 
Q1. The correct option(s) related to adsorption processes is(are) 
A. Chemisorption results in a unimolecular layer. 
B. The enthalpy change during physisorption is in the range of 100 to ?????? ???? ?? ???? - ?? . 
C. Chemisorption is an endothermic process. 
D. Lowering the temperature favors physisorption processes. 
Ans: (A,D) 
(A) In Chemisorption unimolecular layer is formed. 
(B) In Phesisorption 20-40 kJ of heat is released. 
(C) Chemisorption is exothermic process. 
(D) Physisorption increases with decrease in temperature. 
 
 
2021 
JEE Advanced 2021 Paper 1 Online 
Q3. The correct statement(s) related to colloids is (are) 
A. The process of precipitating colloidal sol by an electrolyte is called peptization. 
B. Colloidal solution freezes at higher temperature than the true solution at the same 
concentration. 
 
C. Surfactants form micelle above critical micelle concentration (CMC). CMC depends on 
temperature. 
D Micelles are macromolecular colloids. 
Ans: (B,C) 
Statements (b) and (c) are correct whereas statements (a) and (d) are incorrect. 
(a) Process of precipitating colloidal sol is known as coagulation. temperature. 
(c) Micelles are formed at the critical micelle concentration (CMC) which depends to temperature. 
(d) Micelles and macromolecular colloids are two different types of colloid.