CBSE Class 12 Chemistry Surface Chemistry Assignment Set 01

Question. The adsorption of a gas on a solid surface varies with pressure of the gas in which of the following manner
(a) fast → slow → independent of the pressure
(b) slow → fast → independent of the pressure
(c) independent of the pressure → fast → slow
(d) independent of the pressure → slow → fast
Answer : A

Question. 1g of charcoal adsorbs 100 mL 0.5 M CH₃COOH to form a monolayer, and thereby the molarity of CH₃COOH reduces to 0.49. Calculate the surface area of the charcoal covered by each molecule of acetic acid. Surface area of charcoal = 3.01 × 10² m²/g.
(a) 0.5 × 10^–19 m²
(b) 4 × 10–20 m²
(c) 0.5 × 10–10 m²
(d) 5 × 10^–19 m²
Answer : D

Question. 3.6 gram of oxygen is adsorbed on 1.2 g of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 273 K?
(a) 0.19 Lg^–1
(b) 1 Lg^–1
(c) 2.1 Lg^–1
(d) None of these
Answer : C

Question. Flocculation value of BaCl₂ is much less than that of KCl for sol A and flocculation value of Na₂SO₄ is much less than that of NaBr for sol B.
The correct statement among the following is
(a) Both the sols A and B are negatively charged.
(b) Sol A is positively charged and Sol B is negatively charged.
(c) Both the sols A and B are positively charged.
(d) Sol A is negatively charged and sol B is positively charged.
Answer : B

Question. Consider the following statements about Langmuir isotherm.
I. The free gas and adsorbed gas are in dynamic equilibrium.
II. All adsorption sites are equivalent.
III. The initially adsorbed layer can act as a substrate for further adsorption
IV. The ability of a molecule to get adsorbed at a given site is independent of the occupation of neighboring sites.
The correct statements are
(a) I, II, III and IV
(b) I, II and IV
(c) I, III and IV
(d) I, II and III
Answer : B

Question. Freshly obtained SnO₂ is added to water containing a little HCl. The sol obtained would be ................ due to preferential adsorbtion of ................... ions
(a) positively charged, H⁺
(b) negatively charged, SnO₃^2–
(c) positively charged, Sn⁴⁺
(d) positively charged, Sn²⁺
Answer : C

Question. One desires to prepare a positively charged sol of silver iodide. This can be achieved by
(a) adding a little AgNO₃ solution to KI solution in slight excess
(b) adding a little KI solution to AgNO₃ solution in slight excess
(c) mixing equal volumes of equimolar solutions of AgNO₃ and KI
(d) None of these
Answer : B

Question. The curve that best describes the adsorption of a gas (X g) on 1.0 g of a solid substrate as a function of pressure (p) at a fixed temperature is

(a) 1
(b) 2
(c) 3
(d) 4
Answer : B

Question. Which one of the following is an example of homogeneous catalysis ?
(a) Haber’s process of synthesis of ammonia 
(b) Catalytic conversion of SO₂ to SO₃ in contact process
(c) Catalytic hydrogenation of oils
(d) Acid hydrolysis of methyl acetate
Answer : D

Question. Which of the following statements about a catalyst is/are true ?
(a) A catalyst accelerates the reaction by bringing down the free energy of activation
(b) A catalyst also takes part in the reaction mechanism
(c) A catalyst makes the reaction more feasible by making the ΔGº more negative
(d) A catalyst makes the equilibrium constant of the reaction more favourable for the forward reaction.
Answer : A

Question. 100 mL each of two sols of AgI, one obtained by adding AgNO₃ to slight excess of KI and another obtained by adding KI to slight excess of AgNO₃, are mixed together. Then
(a) The two sols will stabilize each other
(b) The sol particles will acquire more electric charge
(c) The sols will coagulate each other mutually
(d) A true solution will be obtained
Answer : C

Question. Under the influence of an electric field, the particles in a sol migrate towards cathode. The coagulation of the same sol is studied using NaCl, Na₂SO₄ and Na₃PO₄ solutions. Their coagulating values will be in the order
(a) NaCl > Na₂SO₄ > Na₃PO₄
(b) Na₂SO₄ > Na₃PO₄ > NaCl
(c) Na₃PO₄ > Na₂SO₄ > NaCl
(d) Na₂SO₄ > NaCl > Na₃PO₄
Answer : A

Question. Which of the following is true in respect of chemical adsorption?
(a) ΔH < 0, ΔS > 0, ΔG > 0
(b) ΔH < 0, ΔS < 0, ΔG < 0
(c) ΔH > 0, ΔS > 0, ΔG < 0
(d) ΔH > 0, ΔS < 0, ΔG > 0
Answer : B

Question. In an experiment, 200 mL of 0.5 M oxalic acid is shaken with 10 g of activated charcoal and filtered. The concentration of the filtrate is reduced to 0.4 M. The amount of adsorption (x/m) is
(a) 0.9
(b) 1.8
(c) 0.18
(d) 0.09
Answer : C

Question. For a gas-solid adsorbent system, the adsorption isotherm x/m = Kp^1/n is applicable, where n = 5 in the specified case. Point out the correct statement :
(a) (x/m)Vs.p plot is a linear graph with slope equal to 5
(b) (x/m)Vs.log x Vs.p log plot is a straight line with slope equal to 5
(c) log(x/m) x Vs. log p is a straight line with slope equal to 0.2
(d) none of these
Answer : C

Question. In an experiment, addition of 4.0 mL of 0.005M BaCl₂ to 16.0 ml of arsenius sulphide sol just causes the complete coagulation in 2 hrs. The flocculating value of the effective ion is
(a) Cl^–, 1.0
(b) Cl^–, 2.0
(c) Ba²⁺, 1.0
(d) Ba²⁺, 0.5
Answer : C

Question. The most appropriate method of making egg– albumin sol is :
(a) Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well.
(b) Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the yellow part of the content to 100 mL of 5% w/V saline solution and homogenize with a mechanical shaker.
(c) Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the white part of the content to 100 mL of 5% w/V saline solution and homogenize with a mechanical shaker
(d) Break an egg carefully and transfer only the yellow part of the content to 100 mL of 5% w/V saline solution and stir well.
Answer : A

Question. To an emulsion of olive oil and water, stabilized by soap of alkali metals, a small amount of salts of bivalent or tervalent cations is added. Then
(a) stabilization of emulsion enhances
(b) the emulsion breaks into the component liquids
(c) reversal of phase to W/O type takes place
(d) the emulsion turns into a true solution
Answer : C

Question. The density of gold is 19 g/cm3. If 1.9 × 10^–4 g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10 nm, then the number of gold particles per mm³ of the sol will be :
(a) 1.9 × 10¹²
(b) 6.3 × 10¹⁴
(c) 6.3 × 10¹⁰
(d) 2.4 × 10⁶
Answer : D

Question. The volume of a colloidal particle, V_C as compared to the volume of a solute particle in a true solution V_S , could be
(a) V_C/V_S ≃ 10³
(b) V_C/V_S ≃ 10^-3
(c) V_C/V_S ≃ 10²³
(d) V_C/V_S ≃ 1
Answer : A

Numeric Value Answer

Question. In an adsorption experiment, a graph between log (x/m) versus log P was found to be linear with a slope of 45°. The intercept on the y axis was found to be 0.301. The amount of the gas adsorbed per gram of charcoal under a pressure of 3.0 atm is ________.
Answer : 6

Question. Considering Freundlich adsorption isotherm in this plot, find the extent of adsorption at a pressure of 4 atm.

Answer : 8

Question. The density of gold is 19 g/cm³. 1.9 × 10^–3 g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10.5 nm.
The gold particles are occupying 50% of the available space. The number of gold particles per mm3 is about 10ⁿ. Find the value of n.
Answer : 7

Question. 10% sites of catalyst bed have adsorbed by H₂. On heating H₂ gas is evolved from sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm³. No. of sites available is 5.4 × 10¹⁶ per cm² and surface area is 1000 cm². Find out the no. of surface sites occupied per molecule of H₂.
(Given NA = 6 × 10²³)
Answer : 3

Question. The coagulation of 400 mL of a positive colloid took place when 1.46 g of HCl was added to it without changing the volume much. What is the flocculation value of HCl for the colloid?
Answer : 100

Question. A sample of 16 g charcoal was brought into contact with CH₄ gas contained in a vessel of 1 litre at 27 °C. The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6 g/cm3. What is the volume of the CH₄ gas adsorbed per gram of the adsorbent at 608 torr and 27 °C?
Answer : 16.25

Question. A cationic colloidal electrolyte forms micelle at 10^–3 M concentration in water. Its 1 mm³ solution contains 2 × 10¹³ micelle structures. What is the number of cations contained in one micelle? Take NA = 6 × 10^23.
Answer : 30

Question. One gram of activated carbon has a surface area of 1000 m². Considering complete coverage as well as monomolecular adsorption, how much ammonia at 1 atm and 273 K would be adsorbed on the surface of 44/7 g carbon if radius of a ammonia molecules is 10^–8 cm.
Answer : 7.46

Question. The coagulation of 100 mL of a colloidal sol of gold is completely prevented by addition of 250 mg of a substance to it before adding 10 mL of 1% NaCl solution. What is the gold number of this substance?
Answer : 25

Question. A detergent (C₁₂H₂₅SO₄^-Na⁺ ) solution becomes a colloidal sol at a concentration of 10^–3 M. On an average 10¹³ colloidal particles are present in 1 mm₃. What is the average number of ions which are contained by onecolloidal particle (micelle)?
Answer : 60
 

LEVEL-1 QUESTIONS

Question 1. What do you mean by Electrophoresis?
Answer: Electrophoresis is the movement of colloidal particles towards the cathode or anode under the influence of an electric field.
In simple words: Charged colloidal particles move towards oppositely charged electrodes when electricity is passed through the solution.

Exam Tip: Mention the movement of charged particles toward oppositely charged electrodes to secure full marks.

Question 2. Describe cleansing action of soaps and detergents.
Answer: Soaps and detergents contain a hydrophobic tail (hydrocarbon chain) that dissolves in grease or oil and a hydrophilic head (ionic part) that dissolves in water. When soap is added to oily fabric, the soap molecules form spherical aggregates called micelles. The hydrophobic part traps the grease inside, while the hydrophilic part remains in water, allowing the oil droplet to be easily rinsed away during washing.
In simple words: Soap molecules form clusters called micelles that trap grease inside and wash it away with water.

Exam Tip: Key terms like micelle formation, hydrophobic tail, and hydrophilic head must be included in your explanation.

Question 3. Explain what is observed when:
i) A beam of light is passed through a colloidal solution.
ii) An electric current is passed through a colloidal solution.
Answer:
i) Scattering of light takes place, which is known as the Tyndall Effect, making the path of the light beam visible.
ii) Movement of charged particles takes place towards the oppositely charged electrodes, a process known as electrophoresis.
In simple words: Light gets scattered by colloidal particles, and electricity causes charged particles to move towards electrodes.

Exam Tip: Clearly link phenomenon (i) to Tyndall effect and phenomenon (ii) to electrophoresis.

Question 4. Write short notes on:
i) Brownian movement.
ii) Hardy and Schulze rule.
Answer:
i) Brownian movement: It is the continuous, zig-zag random motion of colloidal particles suspended in a liquid medium, caused by the unbalanced bombardment of molecules of the dispersion medium on the colloidal particles.
ii) Hardy-Schulze rule: According to this rule, oppositely charged ions are effective in causing coagulation of a sol. The coagulating power of an effective ion increases with the valency of the coagulating ion.
In simple words: Brownian movement is random zig-zag motion, and Hardy-Schulze rule states that higher valency ions coagulate colloids faster.

Exam Tip: State clearly that coagulating power is directly proportional to the valency of the active ion.

Question 5. What is difference between Physisorption and chemisorption?
Answer:
Physisorption: It is multi-layered, held by weak Van der Waals forces, has low enthalpy of adsorption, is reversible, and is non-specific in nature.
Chemisorption: It is uni-layered, involves chemical bond formation, has high enthalpy of adsorption, is highly specific, and is irreversible.

In simple words: Physisorption is weak and reversible with multiple layers, while chemisorption is strong and irreversible with a single chemical layer.

Exam Tip: Tabulate the differences based on nature of forces, enthalpy, reversibility, and specificity for clarity.

Question 6. What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Answer: An adsorption isotherm is a curve that depicts the variation in the amount of gas adsorbed by an adsorbent with pressure at a constant temperature. Freundlich Adsorption Isotherm gives an empirical relationship between the quantity of gas adsorbed by mass of adsorbent and pressure at a particular temperature:
\(\frac{x}{m} = kp^{1/n}\)
Where \(x\) is the mass of the adsorbate, \(m\) is the mass of the adsorbent, \(p\) is the pressure, and \(k\) and \(n\) are constants which depend on the nature of adsorbent and gas at a particular temperature.
In simple words: An adsorption isotherm is a graph showing how adsorption changes with pressure, represented by the Freundlich equation.

Exam Tip: Define every term in the formula (\(x/m\), \(p\), \(k\), \(n\)) when writing the Freundlich equation.

Question 7. What are lyophilic and lyophobic sols? Give one example of each type.
Answer:
Lyophilic sols: These are solvent-loving sols, stable, reversible, and can be prepared directly by mixing dispersed phase with dispersion medium. Example: Starch sol.
Lyophobic sols: These are solvent-hating sols, unstable, irreversible, and must be prepared by special chemical or physical methods. Example: Gold sol. Hydrophobic sols are easily coagulated due to less affinity towards the dispersion medium.
In simple words: Lyophilic sols love the solvent and are stable, while lyophobic sols hate the solvent and are unstable.

Exam Tip: Mention stability and preparation methods as key differentiating points between lyophilic and lyophobic sols.

LEVEL-2 QUESTIONS

Question 1. Explain:
i) Sun looks red at the time of sun set.
ii) Physisorption is multimolecular while chemisorption is monomolecular.
Answer:
i) At sunset, sunlight travels a relatively longer distance through the atmosphere. As a consequence, shorter wavelengths (blue part) are scattered away by atmospheric particles, leaving the longer wavelengths (red part) to reach our eyes, making the sun look red.
ii) Chemisorption occurs as a result of chemical bond formation between adsorbent and adsorbate. Once the surface of the adsorbent is covered with one chemical layer, no further chemical reaction can take place, making it monomolecular. In contrast, physisorption is held by weak Van der Waals forces, so any number of layers may be formed one over the other.
In simple words: Sun looks red because blue light scatters away over long distances, and physisorption forms multiple layers unlike single-layer chemisorption.

Exam Tip: Use the concept of wavelength scattering for sunset and chemical bonding valency for adsorption layers.

Question 2. Explain what is observed when:
i) An electrolyte, KCl is added to hydrated ferric oxide sol.
ii) An electric current is passed through a colloidal solution.
iii) A beam of strong light is passed through a colloidal solution.
Answer:
i) Coagulation occurs because ferric hydroxide sol is positively charged and gets neutralized by oppositely charged chloride ions of the added electrolyte.
ii) Movement of colloidal particles takes place towards the oppositely charged electrodes (electrophoresis).
iii) Light scattering takes place, also known as the Tyndall effect.
In simple words: Adding salt causes coagulation, electricity causes particle movement, and light beam causes scattering (Tyndall effect).

Exam Tip: Mention charge neutralization for electrolyte coagulation and light scattering for Tyndall effect.

Question 3. What is difference between multimolecular and macromolecular colloids? Give one example of each.
Answer:
Multimolecular colloids: They contain dispersed particles formed by the aggregation of a large number of atoms or molecules having diameters of less than 1 nm. These are usually lyophobic sols and particles are held by weak Van der Waals forces. Example: \(As_2S_3\), Gold sol.
Macromolecular colloids: They are molecularly dissolved solutions of polymers with particle size in the colloidal range and are lyophilic in nature. Particles are held by chemical bonds. Example: Starch, cellulose, nylon.
In simple words: Multimolecular colloids consist of small aggregated molecules, while macromolecular colloids consist of giant polymer molecules.

Exam Tip: Focus on particle formation (aggregation vs large polymer molecules) and bonding forces.

Question 4. What is meant by term peptizaion?
Answer: Peptization is the process of converting a freshly prepared precipitate into colloidal particles by adding a suitable electrolyte (peptizing agent).
In simple words: Peptization breaks down a precipitate into a colloidal solution by adding an electrolyte.

Exam Tip: Always mention 'freshly prepared precipitate' when defining peptization.

Question 5. Explain why hydrophobic sols are relatively more stable than hydrophobic sols. [Note: Question as transcribed from source material comparing lyophilic and hydrophobic stability contexts]
Answer: Lyophilic sols are much more stable than hydrophobic sols because lyophilic colloids have a strong affinity and solvation layer with the dispersion medium, protecting the particles from aggregation.
In simple words: Lyophilic sols have greater affinity towards the dispersion medium, making them harder to coagulate than hydrophobic sols.

Exam Tip: Emphasize the strong interaction and solvation shell around lyophilic particles.

Question 6. What are emulsifiers?
Answer: Emulsifiers are stabilizing agents that are added to emulsions to stabilize them and prevent the separation of the immiscible phases.
In simple words: Emulsifiers are substances that keep emulsions stable and mixed.

Exam Tip: State their role in reducing interfacial tension between oil and water phases.

Question 7. Give reason why a finely divided substance is more effective as an adsorbent.
Answer: A finely divided substance has a much larger surface area per unit mass, which provides more active sites for adsorption, making it more effective.
In simple words: Smaller particles have more surface area, providing more space for adsorption.

Exam Tip: State clearly that adsorption is a surface phenomenon directly proportional to surface area.

Question 8. What are zeolites? Describe some of their features. Give two applications of zeolites.
Answer: Zeolites are microporous, aluminosilicate minerals with a honeycomb-like structure. They act as shape-selective catalysts because their catalytic action depends upon the pore size of the catalyst and the size of reactant and product molecules.
Applications:
1. ZSM-5 is used to convert alcohols directly into gasoline.
2. Used in softening of hard water through ion exchange.
In simple words: Zeolites are shape-selective porous catalysts used in petroleum cracking and water softening.

Exam Tip: Mention 'shape-selective catalysis' and the specific example of ZSM-5 for petroleum refining.

LEVEL-3 QUESTIONS

Question 1. What is meant by Autocatalysis and Induced catalysis?
Answer:
Autocatalysis: It is a phenomenon in which one of the products formed during a chemical reaction acts as a catalyst and speeds up the reaction. For example, during the oxidation of oxalic acid by acidified \(KMnO_4\), \(Mn^{2+}\) ions are produced which act as a catalyst and speed up the reaction as it progresses.
Induced catalysis: When one chemical reaction influences the rate of another reaction that does not occur on its own under similar conditions, it is called induced catalysis.
In simple words: Autocatalysis happens when a reaction product acts as its own catalyst, while induced catalysis occurs when one reaction accelerates another.

Exam Tip: Quote the classic example of \(Mn^{2+}\) autocatalysis in oxalic acid titration with \(KMnO_4\).

Question 2. Answer the following:
i) Why silica gel is used as dehumidizer?
ii) What is the significance of a gold number?
Answer:
i) Silica gel is a strong adsorber of moisture present in air due to its porous structure and high surface area, which is why it is used as a dehumidifier.
ii) Gold number is a measure of the protective power of lyophilic colloids. Smaller the value of gold number, greater will be the protective power of the colloid.
In simple words: Silica gel absorbs moisture to keep things dry, and a lower gold number means better protective power for colloids.

Exam Tip: Remember that gold number is inversely proportional to protective power.

Question 3. Explain:
i) Persistent dialysis of a colloidal sol is carried out.
ii) Gelatin is added to gold sol.
Answer:
i) A small amount of electrolyte is required for the stability of a colloidal sol. On persistent dialysis, the electrolyte is almost completely removed, which destabilizes and leads to coagulation of the sol.
ii) Gelatin acts as a protective lyophilic colloid. When added to gold sol (a lyophobic sol), it forms a protective layer around gold particles, increasing its stability.
In simple words: Removing all electrolytes by dialysis causes coagulation, and adding gelatin protects lyophobic gold sol.

Exam Tip: Highlight that traces of electrolytes are essential for sol stability.

Question 4. A colloidal solution of AgI is prepared by 2 different methods-
a) Addition of silver nitrate in KI
b) Addition of KI in AgNO_3
i) What is the charge AgI colloidal particles in 2 test tubes A & B.
ii) Give reason    for the origin of the charge
iii) [Part iii transcribed from source: explanation of test tube charges]
Answer:
i) Test tube A (Addition of \(AgNO_3\) in excess KI) has negatively charged colloidal particles (\(AgI/I^-\)), while test tube B (Addition of KI in excess \(AgNO_3\)) has positively charged colloidal particles (\(AgI/Ag^+\)).
ii) The origin of the charge is due to preferential adsorption of common ions from the dispersion medium onto the surface of the colloidal particles.
iii) Test tube A has negative charge while test tube B has positive charge on colloidal particles due to preferential adsorption.
In simple words: The charge on colloidal AgI depends on which reactant is in excess due to preferential adsorption of ions.

Exam Tip: Always apply the rule of preferential adsorption of common ions to determine surface charge in precipitation reactions.