Maharashtra Board Class 11 Chemistry Chapter 11 Adsorption and Colloids Solutions

11th Chemistry Chapter 11 Exercise Adsorption And Colloids Solutions Maharashtra Board

Adsorption And Colloids Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 11 Exercise Solutions

1. Choose The Correct Option.

Question A. The size of colloidal particles lies between
(a) 10-10 m and 10-9 m
(b) 10-9 m and 10-6 m
(c) 10-6 m and 10-4 m
(d) 10-5 m and 10-2 m
Answer: (b) 10-9 m and 10-6 m
In simple words: Colloidal particles are larger than true solution particles but smaller than suspension particles, fitting within the nanometer to micrometer range.

🎯 Exam Tip: Remember the specific size range of colloidal particles as it's a fundamental characteristic that distinguishes them from true solutions and suspensions.

Question B. Gum in water is an example of
(a) true solution
(b) suspension
(c) lyophilic sol
(d) lyophobic sol
Answer: (c) lyophilic sol
In simple words: Gum in water is a lyophilic sol because it readily forms a colloidal dispersion when mixed with water due to strong affinity.

🎯 Exam Tip: Distinguishing between lyophilic and lyophobic sols based on common examples is crucial for understanding their properties.

Question C. In Haber process of production of ammonia K2O is used as
(a) catalyst
(b) inhibitor
(c) promotor
(d) adsorbate
Answer: (c) promotor
In simple words: K2O acts as a promotor in the Haber process, enhancing the activity of the main catalyst (iron) for ammonia production.

🎯 Exam Tip: Knowledge of specific additives like promotors and inhibitors in industrial processes, such as the Haber process, is important for applied chemistry questions.

Question D. Fruit Jam is an example of-
(a) sol
(b) gel
(c) emulsion
(d) true solution
Answer: (b) gel
In simple words: Fruit jam is a gel, which is a colloidal system where a liquid is dispersed in a solid medium, giving it a semi-solid consistency.

🎯 Exam Tip: Classifying everyday substances into different types of colloidal systems (sol, gel, emulsion) helps solidify understanding of colloid chemistry.

2. Answer In One Sentence :

Question A. Name type of adsorption in which van der Waals focres are present.
Answer: Physical adsorption or physisorption.
In simple words: Physisorption is the type of adsorption driven by weak intermolecular forces like van der Waals forces.

🎯 Exam Tip: Knowing the types of forces involved is key to differentiating between physical and chemical adsorption.

Question B. Name type of adsorption in which compound is formed.
Answer: Chemical adsoiption or chemisorption.
In simple words: Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent, leading to a new surface compound.

🎯 Exam Tip: Differentiating physisorption and chemisorption based on bond formation is a critical concept often tested.

Question C. Write an equation for Freundlich adsorption isotherm.
Answer: Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\( \frac{x}{m} = k P^{1/n} \left( n > 1 \right) \).......(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\( \frac{x}{m} \) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
In simple words: The Freundlich isotherm equation relates the amount of gas adsorbed per unit mass of adsorbent to the equilibrium pressure, using constants that reflect the specific adsorption system.

🎯 Exam Tip: Be able to write and explain the terms in the Freundlich adsorption isotherm equation, including the conditions for the constant 'n'.

3. Answer The Following Questions:

Question A. Define the terms:
(a) Inhibition
(b) Electrophoresis
(c) Catalysis.
Answer:
(a) Inhibition:
The phenomenon in which the rate of chemical reaction is reduced by an inhibitor is called inhibition.
(b) Electrophoresis:
The movement of colloidal particles under an applied electric potential is called electrophoresis.
(c) Catalysis:
The phenomenon of increasing the rate of a chemical reaction with the help of a catalyst is known as catalysis.
In simple words: Inhibition slows down reactions, electrophoresis moves charged colloidal particles with electricity, and catalysis speeds up reactions using catalysts.

🎯 Exam Tip: Clear and concise definitions for these terms are essential, especially distinguishing between inhibition and catalysis which have opposite effects on reaction rates.

Question B. Define adsorption. Why students can read blackboard written by chalks?
Answer:
• Adsorption is the phenomenon of accumulation of higher concentration of 'one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
• When we write on blackboard using chalk, the chalk particles get adsorbed on the surface of the blackboard.
Hence, students can read blackboard written by chalks.
In simple words: Adsorption is the surface accumulation of one substance on another due to unbalanced forces; students can read chalk on a blackboard because chalk particles adsorb onto the blackboard's surface.

🎯 Exam Tip: Understanding adsorption as a surface phenomenon and providing a relevant example like chalk on a blackboard demonstrates a clear grasp of the concept.

Question C. Write characteristics of adsorption.
Answer: Following are the characteristics of adsorption:
• Adsorption is a surface phenomenon.
• It depends upon the surface area of the adsorbent.
• It involves physical forces (van der Waals forces) or chemical forces (chemical or covalent bonds).
• Adsorbate is always present in higher concentration on the surface of an adsorbent than in the bulk.
• Adsorption is dependent on temperature (of the surface) and pressure (of adsorbate gas).
• It takes place with the evolution of heat (with some exceptions).
In simple words: Adsorption is a surface process influenced by surface area, temperature, and pressure, involving either weak physical or strong chemical forces, and usually releases heat.

🎯 Exam Tip: Listing multiple characteristics of adsorption, covering its physical nature, factors affecting it, and energy changes, is good for comprehensive answers.

Question D. Distinguish between Lyophobic and Lyophilic sols.
Answer:
Lyophobic sols (colloids):
1. Lyophobic sols are formed only by special methods.
2. They are irreversible.
3. These are unstable and hence, require traces of stabilizers.
4. Addition of small amount of electrolytes causes precipitation or coagulation of lyophobic sols.
5. Viscosity of lyophobic sol is nearly the same as the dispersion medium.
6. Surface tension of lyophobic sol is nearly the same as the dispersion medium.
Lyophilic sols (colloids):
1. Lyophilic sols are formed easily by direct mixing.
2. They are reversible.
3. These are self-stabilized.
4. Addition of large amount of electrolytes causes precipitation or coagulation of lyophilic sols.
5. Viscosity of lyophilic sol is much higher than that of the dispersion medium.
6. Surface tension of lyophilic sol is lower than that of dispersion medium.
In simple words: Lyophobic sols are difficult to prepare, unstable, and irreversible with low affinity for the medium, while lyophilic sols are easily formed, stable, reversible, and have a high affinity for the medium.

🎯 Exam Tip: When distinguishing between two concepts, use a point-by-point comparison to highlight clear differences in their properties and formation.

Question E. Identify dispersed phase and dispersion medium in the following colloidal dispersions.
(a) milk
(b) blood
(c) printing ink
(d) fog
Answer:

🎯 Exam Tip: Practice identifying the dispersed phase and dispersion medium for various everyday colloidal systems to understand their classification.

Question F. Write notes on :
(a) Tyndall effect
(b) Brownian motion
(c) Types of emulsion
(d) Hardy-Schulze rule
Answer:
(a) Tyndall effect:
i. Tyndall observed that when light passes through true solution, the path of light through it cannot be detected.
ii. However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions and the path of the light through colloidal dispersion becomes visible to observer standing at right angles to its path.
iii. The phenomenon of scattering of light by colloidal particles and making path of light visible through the dispersion is referred as Tyndall effect and the bright cone of the light is called Tyndall cone.
iv. Tyndall effect is observed only when the following conditions are satisfied.
• The diameter of the dispersed particles is not much smaller than the wavelength of light used.
• The refractive indices of dispersed phase and dispersion medium differ largely.
v. Significance of Tyndall effect:
• It is useful in determining number of particles in colloidal system and their particle size.
• It is used to distinguish between colloidal dispersion and true solution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र टिंडल प्रभाव को दर्शाता है। इसमें एक प्रकाश स्रोत से निकली किरण एक विलयन और एक कोलाइडल परिक्षेपण से गुजरती है। विलयन में प्रकाश का मार्ग अदृश्य रहता है, जबकि कोलाइडल परिक्षेपण में, प्रकाश का मार्ग एक चमकदार शंकु (टिंडल शंकु) के रूप में दिखाई देता है, जिससे प्रकाश का प्रकीर्णन स्पष्ट होता है।
(b) Brownian motion:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:
• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.
(c) Types of emulsion:
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.
(d) Hardy-Schulze rule:
i. Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.
ii. In the coagulation of negative sol, the flocculating power follows the following order:
\( \text{Al}^{3+} > \text{Ba}^{2+} > \text{Na}^{+} \)
iii. Similarly, in the coagulation of positive sol, the flocculating power is in the following order:
\( [\text{Fe}(\text{CN})_6]^{4-} > \text{PO}_4^{3-} > \text{SO}_4^{2-} > \text{Cl}^{-} \)
In simple words: The Tyndall effect explains light scattering by colloids, Brownian motion is the random movement of colloidal particles, emulsions are liquid-liquid colloids (oil-in-water or water-in-oil), and the Hardy-Schulze rule describes how ion valency affects colloid coagulation.

🎯 Exam Tip: For "write notes on" questions, ensure each point is clearly defined, with causes/conditions and examples where applicable, especially for the Tyndall effect and Brownian motion.

Question G. Explain Electrophoresis in brief with the help of diagram. What are its applications ?
Answer:
i. Electrophoresis: Electrophoresis set up is shown in the diagram below.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र वैद्युतकणसंचलन (इलेक्ट्रोफोरेसिस) का सेटअप दर्शाता है। एक U-ट्यूब में कोलाइडल विलयन भरा है, जिसमें कैथोड और एनोड प्लैटिनम इलेक्ट्रोड डूबे हुए हैं। जब विद्युत विभव लगाया जाता है, तो कोलाइडल कण एक इलेक्ट्रोड की ओर बढ़ते हैं और जमा हो जाते हैं, जबकि विलयन ऊपर जलाशय में रहता है।
• The diagram shows U tube set up in which two platinum electrodes are dipped in a colloidal solution.
• When electric potential is applied across two electrodes, colloidal particles move towards one or other electrode.
• The movement of colloidal particles under an applied electric potential is called electrophoresis.
• Positively charged particles move towards cathode while negatively charged particles migrate towards anode and get deposited on the respective electrode.
ii. Applications of electrophoresis:
• On the basis of direction of movement of the colloidal particles under the influence of electric field, it is possible to know the sign of charge on the particles.
• It is also used to measure the rate of migration of sol particles.
• Mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in mixture migrate with different rates.
In simple words: Electrophoresis is the movement of charged colloidal particles in an electric field, used to determine particle charge, measure migration rates, and separate different colloids.

🎯 Exam Tip: When explaining a concept with a diagram, ensure the textual explanation clearly references the diagram's components and processes, and also list its practical applications.

Question H. Explain why finely divided substance is more effective as adsorbent?
Answer:
• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon surface area of the adsorbent.
• Adsorption increases with increase in surface area of the adsorbent.
• Finely divided powdered substances provide larger surface area for a given mass.
Hence, finely divided substance is more effective as adsorbent.
In simple words: Finely divided substances are better adsorbents because adsorption occurs on surfaces, and smaller particles provide a significantly larger total surface area for a given mass, enhancing the adsorption capacity.

🎯 Exam Tip: Always link the effectiveness of an adsorbent directly to its surface area, as adsorption is fundamentally a surface phenomenon.

Question I. What is the adsorption Isotherm?
Answer: The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in case of gas) or concentration (in case of solution) at a given constant temperature is called an adsorption isotherm.
In simple words: An adsorption isotherm is a graph or equation showing how much substance is adsorbed on a surface at a constant temperature, as pressure or concentration changes.

🎯 Exam Tip: Define adsorption isotherm precisely, mentioning the key variables (amount adsorbed, pressure/concentration) and the constant condition (temperature).

Question J. Aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.
Answer:
• When aqueous solution of raw sugar is passed over beds of animal charcoal, charcoal adsorbs the coloured particles from the raw sugar.
• Thus, due to the adsorption of coloured particles, raw sugar becomes colourless when passed over beds of animal charcoal.
In simple words: Animal charcoal removes color from raw sugar solution by adsorbing the colored impurities onto its surface, making the sugar solution colorless.

🎯 Exam Tip: Use practical examples like decolorization of sugar to illustrate the utility and mechanism of adsorption.

Question K. What happens when a beam of light is passed through a colloidal sol?
Answer:
i. When a beam of light is passed through colloidal sol, it is observed that the colloidal particles scatter some of the incident light in all directions.
ii. Because of this scattering of light, the path of light through the colloidal dispersion becomes visible to observer standing at right angles to its path and the phenomenon is known as Tyndall effect.
iii.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र टिंडल प्रभाव को दर्शाता है। इसमें एक प्रकाश स्रोत से निकली किरण एक विलयन और एक कोलाइडल परिक्षेपण से गुजरती है। विलयन में प्रकाश का मार्ग अदृश्य रहता है, जबकि कोलाइडल परिक्षेपण में, प्रकाश का मार्ग एक चमकदार शंकु (टिंडल शंकु) के रूप में दिखाई देता है, जिससे प्रकाश का प्रकीर्णन स्पष्ट होता है।
In simple words: When light passes through a colloidal sol, the particles scatter the light, making the path of the light visible, a phenomenon known as the Tyndall effect.

🎯 Exam Tip: Connect the scattering of light by colloidal particles directly to the visibility of the light path, defining it as the Tyndall effect.

Question L. Mention factors affecting adsorption of gas on solids.
Answer: Adsorption of gases on solids depends upon the following factors:
• Nature of adsorbate (gas)
• Nature of solid adsorbent
• Surface area of adsorbent
• Temperature of the surface
• Pressure of the gas
In simple words: Adsorption of gases on solids is influenced by the type of gas and solid, the adsorbent's surface area, and the temperature and pressure conditions.

🎯 Exam Tip: Listing all relevant factors is crucial for questions asking about parameters affecting a process; ensure you include both adsorbate and adsorbent properties, and environmental conditions.

Question M. Give four uses of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):
• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, \( \text{H}_2\text{SO}_4 \) (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.
ii. Gas masks:
• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.
iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:
• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.
In simple words: Adsorption is used in industrial catalysis to speed up reactions, in gas masks to purify air, to control humidity with desiccants, and to achieve high vacuum by removing residual gases.

🎯 Exam Tip: Provide diverse applications of adsorption, demonstrating its importance in various fields, and give a brief explanation for each use case.

Question N. Explain Bredig's arc method.
Answer:
• Colloidal sols can be prepared by electrical disintegration using Bredig's arc method.
• This process involves vaporization as well as condensation.
• Colloidal sols of metals such as gold, silver, platinum can be prepared by this method.
• In this method, electric arc is struck between electrodes of metal immersed in the dispersion medium.
• The intense heat produced vapourizes the metal which then condenses to form particles of colloidal sol.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ब्रेडीग की आर्क विधि को दर्शाता है, जिसका उपयोग कोलाइडल सॉल बनाने में होता है। इसमें धातु के इलेक्ट्रोड को परिक्षेपण माध्यम (जैसे पानी) में डुबोया जाता है, जिसके चारों ओर बर्फ रखी होती है। इलेक्ट्रोड के बीच एक विद्युत आर्क उत्पन्न होता है, जिससे धातु वाष्पीकृत होती है और फिर संघनित होकर कोलाइडल कण बनाती है।
In simple words: Bredig's arc method prepares metal colloids by generating an electric arc between metal electrodes submerged in a dispersion medium, causing the metal to vaporize and then condense into colloidal particles.

🎯 Exam Tip: Describe the key steps of Bredig's arc method – electrical disintegration, vaporization, and condensation – and mention the types of sols it produces.

Question O. Explain the term emulsions and types of emulsions.
Answer:
i. A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion.
ii. There are liquid-liquid colloidal systems in which both liquids are either completely or partially immiscible.
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.
In simple words: Emulsions are colloidal mixtures of two immiscible liquids, categorized as oil-in-water (like milk) where oil is dispersed in water, or water-in-oil (like butter) where water is dispersed in oil.

🎯 Exam Tip: Clearly define emulsion, distinguish between o/w and w/o types, and provide at least two common examples for each type.

4. Explain The Following :

Question A. A finely divided substance is more effective as adsorbent.
Answer:
• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon the surface area of the adsorbent.
• Adsorption increases with an increase in surface area of the adsorbent.
• Finely divided powdered substances provide a larger surface area for a given mass. Hence, a finely divided substance is more effective as an adsorbent.
In simple words: Finely divided substances are superior adsorbents because they offer a significantly larger total surface area for a given mass, allowing more sites for adsorption to occur.

🎯 Exam Tip: Always emphasize the direct relationship between increased surface area and enhanced adsorption efficiency when explaining this concept.

Question B. Freundlich adsorption isotherm, with the help of a graph.
Answer: Graphical representation of the Freundlich adsorption isotherm:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्रेंडलिच अधिशोषण समतापी के आलेखीय निरूपण को दर्शाता है। इसमें x/m (अधिशोषित गैस का द्रव्यमान प्रति अधिशोषक के द्रव्यमान) को संतुलन दाब (P) के विरुद्ध आलेखित किया गया है। वक्र दर्शाता है कि दाब बढ़ने के साथ अधिशोषण बढ़ता है, लेकिन उच्च दाब पर यह संतृप्त हो जाता है।
i. Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\( \frac{x}{m} = k P^{1/n} \left( n > 1 \right) \)..........(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\( \frac{x}{m} \) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
ii. The graphical representation of Freundlich equation is as shown in the adjacent plot of x/m vs 'P'.
iii. In case of solution, P in the equation (i) is replaced by the concentration (C) and thus,
\( \frac{x}{m} = k C^{1/n} \)..........(ii)
iv. By taking logarithm on both sides of the equation (ii),
we get
\( \log \frac{x}{m} = \log k + \frac{1}{n} \log C \)..........(iii)
v. On plotting a graph of \( \log \frac{x}{m} \) against \( \log C \) or \( \log P \), a straight line is obtained as shown in the adjacent plot. The slope of the straight line is \( \frac{1}{n} \) and intercept on Y-axis is \( \log k \).
vi. The factor \( \frac{1}{n} \) ranges from 0 to 1. Equation (iii) holds good over limited range of pressures.
a. When \( \frac{1}{n} \to 0 \), \( \frac{x}{m} \to \text{constant} \), the adsorption is independent of pressure.
b. When \( \frac{1}{n} = 1 \), \( \frac{x}{m} = k P \), i.e., \( \frac{x}{m} \propto P \), the adsorption varies directly with pressure.
c. The experimental isotherms tend to saturate at high pressure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्रेंडलिच अधिशोषण समतापी के लघुगणकीय रूप का आलेखीय निरूपण दर्शाता है। इसमें \( \log \left( \frac{x}{m} \right) \) को \( \log P \) के विरुद्ध आलेखित किया गया है, जिसके परिणामस्वरूप एक सीधी रेखा प्राप्त होती है। इस रेखा का ढलान \( \frac{1}{n} \) और Y-अक्ष पर अंतःखंड \( \log k \) है, जो समीकरण के मापदंडों को स्पष्ट करता है।
In simple words: The Freundlich adsorption isotherm is an empirical equation and its graphical representation that shows how the amount of gas adsorbed on a solid changes with pressure at a constant temperature, often becoming saturated at high pressures.

🎯 Exam Tip: Understand both the mathematical form of the Freundlich isotherm and its graphical representation, including the significance of the slope and intercept in the logarithmic form.

5. Distinguish Between The Following:

Question A. Adsorption and absorption. Give one example.
Answer:
Adsorption:
• Adsorption is a surface phenomenon as adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent.
• Concentration of the adsorbate is high only at the surface of the adsorbent.
• It is dependent on temperature and pressure.
• It is accompanied by evolution of heat known as heat of adsorption.
In simple words: Adsorption is a surface process where substances accumulate on the surface, while absorption involves uniform distribution throughout the bulk of the material.

🎯 Exam Tip: Clearly differentiate adsorption and absorption by focusing on whether the accumulation occurs on the surface or throughout the bulk of the material.

1. In chemisorption, the forces operating are of chemical nature (covalent or ionic bonds).
2. It is highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
3. The heat of adsorption is high and lies in the range 40-200 kJ mol-1.
4. It is favoured at high temperature, however, the extent of chemical adsorption is lowered at very high temperature due to bond breaking.
5. It is irreversible.
6. Chemisorption forms monomolecular layer of adsorbed particles. e.g. N2 gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.
In simple words: Physisorption involves weak Van der Waals forces, is non-specific, reversible, and has low heat of adsorption, forming multi-molecular layers. Chemisorption involves chemical bonds, is highly specific, irreversible, has high heat of adsorption, and forms mono-molecular layers.

🎯 Exam Tip: Focus on the nature of forces, specificity, reversibility, and the heat of adsorption to differentiate between physisorption and chemisorption effectively.

Question 6. Adsorption is surface phenomenon. Explain.
Answer: Consider a surface of a liquid or a solid.
• The molecular forces at the surface of a liquid are unbalanced or in unsaturation state.
• In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by the close contact with other particles.
• Because of the unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus, the substance accumulates on the surface of solid or liquid i.e., the substance gets adsorbed on the surface.
Hence, adsorption is a surface phenomenon.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अधिशोषण की प्रक्रिया को दर्शाता है। इसमें एक "बल्क अणु" (A) को दिखाया गया है जिसके चारों ओर संतुलित बल हैं, जबकि एक "सतह अणु" (B) में असंतुलित बल होते हैं, जो अधिशोषक की सतह पर अणुओं के जमाव को प्रेरित करते हैं। यह दर्शाता है कि अधिशोषण एक सतही घटना है।
In simple words: Adsorption is a surface phenomenon because surface molecules or ions have unsatisfied attractive forces, causing them to attract and accumulate other substances, concentrating them only on the surface rather than throughout the bulk.

🎯 Exam Tip: Understanding the unbalanced forces at the surface is key to explaining why adsorption occurs primarily on surfaces.

Question 7. Explain how the adsorption of gas on solid varies with
(a) nature of adsorbate and adsorbent
(b) surface area of adsorbent
Answer: i. a. Nature of adsorbate:
1. All solids adsorb gases to some extent. It is observed that gases having high critical temperature liquify easily and can be readily adsorbed.
2. The gases such as SO2, Cl2, NH3 which are easily liquefiable are adsorbed to a larger extent as compared to gases such as N2, O2, H2, etc. which are difficult to liquify.
3. Thus, the amount of gas adsorbed by a solid depends on the nature of the adsorbate gas i.e., whether it is easily liquefiable or not.
b. Nature of adsorbent: Substances which provide large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases. e.g. Silica gel and charcoal are effective adsorbents due to their porous nature.
ii. Surface area of the adsorbent:
• Adsorption is a surface phenomenon. Hence, the extent of adsorption increases with increase in surface area of the adsorbent.
• Finely divided substances, rough surfaces, colloidal substances are good adsorbents as they provide larger surface area for a given mass.
Note: Critical temperature of some gases and volume adsorbed.

🎯 Exam Tip: Remember that easily liquefiable gases and adsorbents with large surface areas are key factors for efficient adsorption, which is directly linked to the extent of adsorption. The critical temperature table is a good example to illustrate this point.

Question 8. Explain two applications of adsorption.
Answer: i. Catalysis (Heterogeneous catalysis):
• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H2SO4 (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.
ii. Gas masks:
• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.
iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:
• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.
In simple words: Adsorption is used in heterogeneous catalysis, where solid catalysts provide surfaces for reactions, and in gas masks, where adsorbents remove poisonous gases. It also helps control humidity using gels and achieves high vacuum by adsorbing residual gases.

🎯 Exam Tip: When listing applications, provide a clear brief explanation for each, ideally including a specific example where applicable, to demonstrate full understanding.

Question 9. Explain micelle formation in soap solution.
Answer:
• Soap molecule has a long hydrophobic hydrocarbon chain called tail which is attached to hydrophilic ionic carboxylate group, called head.
• In water, the soap molecules arrange themselves to form spherical particles that are called micelles.
• In each micelle, the hydrophobic tails of soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere.
• As a result of this, soap dispersion in water is stable.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक साबुन के मिसेल की संरचना को दर्शाता है, जहाँ साबुन के अणु पानी में एक गोलाकार आकृति बनाते हैं। इसमें हाइड्रोफिलिक (जल-प्रेमी) सिरे बाहर की ओर पानी की ओर होते हैं, जबकि हाइड्रोफोबिक (जल-विरोधी) पूंछें मिसेल के केंद्र की ओर होती हैं, जिससे साबुन का घोल स्थिर रहता है।
In simple words: Soap molecules, having a hydrophilic head and a hydrophobic tail, arrange themselves in water to form micelles, spherical structures where tails face inwards and heads face outwards, ensuring stability in the dispersion.

🎯 Exam Tip: Clearly describe the two parts of a soap molecule and how their opposing affinities for water lead to the specific spherical arrangement of a micelle.

Question 10. Draw labelled diagrams of the following :
(a) Tyndall effect
(b) Dialysis
(c) Bredig's arc method
(d) Soap micelle
Answer: a. Tyndall effect:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र टिंडल प्रभाव को दर्शाता है, जहाँ एक प्रकाश स्रोत से निकलने वाली किरण एक वास्तविक विलयन (सॉल्यूशन) से गुजरने पर अदृश्य रहती है, जबकि एक कोलाइडल परिक्षेपण (कोलाइडल डिस्पर्शन) से गुजरने पर प्रकाश का मार्ग (टिंडल शंकु) स्पष्ट रूप से दिखाई देता है। यह कोलाइडल कणों द्वारा प्रकाश के प्रकीर्णण के कारण होता है।
b. Dialysis:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र डायलिसिस प्रक्रिया को दर्शाता है, जिसमें एक डायलइजिंग मेम्ब्रेन (अर्ध-पारगम्य झिल्ली) का उपयोग करके कोलाइडल परिक्षेपण से अशुद्धियों (जैसे घुले हुए इलेक्ट्रोलाइट) को हटाया जाता है। छोटे कण झिल्ली से गुजरकर पानी और इलेक्ट्रोलाइट में मिल जाते हैं, जबकि बड़े कोलाइडल कण अंदर रहते हैं।
c. Bredig's arc method:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ब्रैडिग की आर्क विधि को दर्शाता है, जिसका उपयोग धातु के कोलाइडल सोल बनाने के लिए किया जाता है। इसमें धातु के इलेक्ट्रोड को परिक्षेपण माध्यम (पानी) में डुबोया जाता है और एक तीव्र विद्युत आर्क बनाया जाता है, जिससे धातु वाष्पीकृत होकर कोलाइडल कणों में संघनित हो जाती है।
d. Soap micelle:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक साबुन के मिसेल की संरचना को दर्शाता है, जहाँ साबुन के अणु पानी में एक गोलाकार आकृति बनाते हैं। इसमें हाइड्रोफिलिक (जल-प्रेमी) सिरे बाहर की ओर पानी की ओर होते हैं, जबकि हाइड्रोफोबिक (जल-विरोधी) पूंछें मिसेल के केंद्र की ओर होती हैं, जिससे साबुन का घोल स्थिर रहता है।
In simple words: These are visual representations of key concepts in colloids and surface chemistry: Tyndall effect shows light scattering by colloidal particles, Dialysis illustrates purification of colloids, Bredig's arc method depicts colloidal sol preparation, and a Soap micelle illustrates the self-assembly of amphiphilic molecules in solution.

🎯 Exam Tip: For diagrams, ensure all components are clearly labelled, and the diagram accurately represents the process or structure to score full marks.

Activity :

Question. Collect the information about methods to study surface chemistry.
Answer: Following are the few methods that are employed to study surface chemistry.
i. X-ray photoelectron spectroscopy:
It is a surface-sensitive spectroscopic technique which is used to measure elemental composition of the surface, to determine elements that are present as contaminants on the surface, etc.
ii. Auger electron spectroscopy:
It is a common analytical technique which is used to study surfaces of materials.
iii. Temperature programmed desorption (TPD):
Adsorbed molecules get desorbed when the surface temperature is increased. TPD technique is used to observe these desorbed molecules and helps in providing information about binding energy between the adsorbate and adsorbent.
iv. Scanning Electron Microscopy:
In this technique, a scanning electron microscope is used to focus electron beam over the surface of the sample to be examined. The electron beam interacts with the sample and an image is obtained. This image provides information about surface structure and composition of the sample.
[Note: Students are expected to collect additional information about surface chemistry on their own.]
In simple words: Surface chemistry is studied using techniques like X-ray photoelectron spectroscopy for elemental composition, Auger electron spectroscopy for general surface analysis, Temperature programmed desorption for binding energy information, and Scanning Electron Microscopy for surface structure and composition imaging.

🎯 Exam Tip: When describing methods, briefly explain the principle and the type of information each technique provides regarding surface properties.

Can You Tell? (Textbook Page No. 160)

Question 1. What is adsorption?
Answer: Adsorption is the phenomenon of accumulation of higher concentration of one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
In simple words: Adsorption is the process where molecules of a substance accumulate on the surface of another substance, forming a higher concentration there due to attractive forces.

🎯 Exam Tip: Ensure your definition of adsorption clearly distinguishes it as a surface phenomenon involving higher concentration on the surface.

Try This. (Textbook Page No. 161)

Question 1. Dip a chalk in ink. What do you observe?
Answer: When a chalk is dipped in ink, it is observed that the ink molecules are adsorbed at the surface of chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption.
In simple words: When chalk is dipped in ink, the colored ink particles adsorb on the chalk's surface, making it colored, while the ink's solvent is absorbed into the bulk of the chalk.

🎯 Exam Tip: This observation provides a clear practical example illustrating both adsorption (color on surface) and absorption (solvent in bulk).

Internet My Friend. (Textbook Page No. 172)

Question. Brownian motion
Answer: Students can search relevant videos on YouTube to visualize Brownian motion.
In simple words: Brownian motion refers to the random, zig-zag movement of microscopic particles suspended in a fluid, caused by collisions with the fluid molecules.

🎯 Exam Tip: While visual aids are helpful, be prepared to describe Brownian motion and its cause textually for exams.

Question. Collect information about Brownian motion.
Answer:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:
• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.
In simple words: Brownian motion is the continuous, random zig-zag movement of colloidal particles in a fluid, caused by their constant, unequal bombardment by the faster-moving molecules of the dispersion medium.

🎯 Exam Tip: For Brownian motion, understanding that it results from the collision of dispersed particles with dispersion medium molecules, transferring kinetic energy, is crucial.

Internet My Friend. (Textbook Page No. 172)

Question 1. Collect information about surface chemistry.
Answer:
• Surface or interface represents the boundary which separates two bulk phases. e.g. Boundary between water and its vapour is a liquid-gas interface.
• Certain properties of substances, particularly of solids and liquids, depend upon the nature of the surface.
• An interface usually has a thickness of a few molecules. However, its area depends on the size of the bulk phase particles.
• Commonly considered bulk phases may be pure compounds or solutions.
• A number of important phenomena, namely, dissolution, crystallization, heterogeneous catalysis, electrode processes and corrosion take place at an interface.
• Thus, study of chemistry of surfaces is critical to many applications in industry, analytical investigations and day-to-day activities such as cleaning and softening of water.
• The branch of chemistry which deals with the nature of surfaces and changes occurring on the surfaces is called surface chemistry.
• Study of surfaces requires a rigorously clean surface. An ultra-clean metal surface can be obtained under very high vacuum, of the order of 10-8 to 10-9 pascal.
• Adsorption, catalysis and colloids (such as emulsions and gels) are some of the important aspects of surface chemistry.
[Note: Students are expected to collect additional information about surface chemistry on their own.]
In simple words: Surface chemistry studies the phenomena occurring at interfaces between phases, which are crucial for processes like catalysis, dissolution, and corrosion, impacting various industrial and daily life applications.

🎯 Exam Tip: Focus on identifying surface chemistry as the study of interfaces and enumerating key phenomena or applications that occur at these interfaces.

Activity. (Textbook Page No. 172)

Question 1. Calculate surface area to volume ratio of spherical particle. See how the ratio increases with the reduction of radius of the particle. Plot the ratio against the radius.
Answer: The graph below shows that as the radius of the spherical particle decreases, the surface to volume ratio increases steadily.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ गोलाकार कणों के लिए त्रिज्या के सापेक्ष सतह क्षेत्र-से-आयतन अनुपात में भिन्नता को दर्शाता है। इसमें दिखाया गया है कि जैसे-जैसे कण की त्रिज्या घटती है, उसका सतह क्षेत्र-से-आयतन अनुपात लगातार बढ़ता जाता है, जिसका अर्थ है कि छोटे कणों में अपेक्षाकृत अधिक सतह क्षेत्र होता है।
In simple words: For a spherical particle, the surface area to volume ratio is \( \frac{3}{r} \). As the radius (r) decreases, this ratio increases, meaning smaller particles have a proportionally larger surface area, which is important for phenomena like adsorption.

🎯 Exam Tip: Remember the inverse relationship between particle radius and surface area to volume ratio; this concept is fundamental to understanding the efficiency of surface-dependent processes.