Request a Free Counselling Session from our Expert Mentor

Welcome to Swastik Classes’ NCERT Solutions for Class 12 Chemistry Chapter 5 – Surface Chemistry. This chapter deals with the study of the behavior of matter at the interface of two phases, such as solid-liquid, liquid-gas, or solid-gas.

Our NCERT Solutions for Class 12 Chemistry Chapter 5 aim to provide students with a comprehensive understanding of the concepts covered in the chapter. Our solutions are designed by subject matter experts with vast experience in the field of chemistry. We have provided step-by-step solutions to all the questions in the chapter, making it easy for students to understand and follow the concepts.

Our solutions also include illustrations and diagrams to help students visualize the processes and remember them better. Additionally, we have included sample questions and answers to help students prepare for their exams.

With our NCERT Solutions for Class 12 Chemistry Chapter 5, students can gain a deeper understanding of the different types of adsorption, the factors affecting adsorption, and the applications of surface chemistry in various fields such as catalysis, colloid chemistry, and corrosion. They can also score well in their exams and build a strong foundation in surface chemistry.

NCERT Solutions For Class 12 Chemistry Chapter 5 – PDF Download

Answers for chemistry class 12 chapter 5 Surface Chemistry

Chemistry Class 12 NCERT Solutions Chapter 5 intext questions and answers

Question 5.1:

Write any two characteristics of Chemisorption.

Answer:

1.  Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.

2.  Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.

Question 5.2:

Why does physisorption decrease with the increase of temperature?

Answer:

Physisorption is exothermic in nature. Therefore, in accordance with Le-Chatelier’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.

Question 5.3:

Why are powdered substances more effective adsorbents than their crystalline forms?

Answer:

Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.

Question 5.4:

Why is it necessary to remove CO when ammonia is obtained by Haber’s process?

Answer:

It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process. CO acts as a poison for the catalyst used. 

Question 5.5:

Why is the ester hydrolysis slow in the beginning and becomes faster after some time?

Answer:

Ester hydrolysis can be represented as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6379/NS(INTEXT)_12-11-08_Sonali_12_Chemistry_5_8_html_46f99c9.gif

The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.

Question 5.6:

What is the role of desorption in the process of catalysis?

Answer:

The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface.

Question 5.7:

What modification can you suggest in the Hardy-Schulze law?

Answer:

Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’

This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’

Question 5.8:

Why is it essential to wash the precipitate with water before estimating it quantitatively?

Answer:

When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.

NCERT Class 12 Chemistry Chapter 5 Exercise Solutions

*Question 5.1:

Distinguish between the meaning of the terms adsorption and absorption.

Give one example of each.

Answer:

Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the ‘adsorbate’ and the substance on whose surface the adsorption takes place is called the ‘adsorbent’. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside.

On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid.

Other example to understand the difference can be: NH3 gets adsorbed on the charcoal where as NH3 when comes in contact with H2O gets absorbed by forming NH4OH solution of uniform concentration.

Question 5.2:

What is the difference between physisorption and chemisorption?

Answer:

Physisorption Chemisorption
1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction. In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.
2. No new compound is formed in the process. New compounds are formed at the surface of the adsorbent.
3. It is generally found to be reversible in nature. It is usually irreversible in nature.
4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol−1. Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol−1.
5. It is favoured by low temperature conditions. It is favoured by high temperature conditions.
6. It is an example of multi-layer adsorption It is an example of mono-layer adsorption.

Question 5.3:

Give reason why a finely divided substance is more effective as an adsorbent.

Answer:

Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.

*Question 5.4:

What are the factors which influence the adsorption of a gas on a solid?

Answer:

There are various factors that affect the rate of adsorption of a gas on a solid surface.

(1) Nature of the gas:

Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because Van der Waal’s forces are stronger in easily liquefiable gases.

(2) Surface area of the solid

The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.

(3) Effect of pressure

Adsorption is a reversible process and is accompanied by a decrease in pressure because gases change their phase. Therefore, adsorption increases with an increase in pressure.

(4) Effect of temperature

Adsorption is an exothermic process. Thus, in accordance with Le- Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

*Question 5.5:

What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_29ca869.jpg

The plot between the extent of adsorption  against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorption isotherm:

Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.

From the given plot it is clear that at pressure PS,  reaches the maximum valve. Ps is called the saturation pressure. Three cases arise from the graph now.

Case I- At low pressure:

The plot is straight and sloping, indicating that the pressure in directly proportional to  i.e., 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_7efff6bf.gif

Case II- At high pressure:

When pressure exceeds the saturated pressure,  becomes independent of P values.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_mbb1deb8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_m6d35ac48.gif

Case III- At intermediate pressure:

At intermediate pressure,  depends on P raised to the powers between 0 and 1. This relationship is known as the Freundlich adsorption isotherm.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_6861fce4.gif

On plotting the graph between log  and log P, a straight line is obtained with the slope equal to  and the intercept equal to log k.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6401/NS_11-11-08_Sonali_12_Chemistry_5_27_html_7d86e422.jpg

Question 5.6:

What do you understand by activation of adsorbent? How is it achieved?

Answer:

By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

(i)  By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.

(ii)  Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.

*Question 5.7:

What role does adsorption play in heterogeneous catalysis?

Answer:

Heterogeneous catalysis:

A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

(i)  Adsorption of reactant molecules on the catalyst surface.

(ii)  Occurrence of a chemical reaction through the formation of an intermediate.

(iii)  De-sorption of products from the catalyst surface

(iv)  Diffusion of products away from the catalyst surface.

In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Question 5.8:

Why is adsorption always exothermic?

Answer:

Adsorption is always exothermic. This statement can be explained in two ways.

(i)  Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

(ii)  ΔH of adsorption is always negative: When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔS is negative. Now for a process to be spontaneous, ΔG should be negative.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6404/NS_11-11-08_Sonali_12_Chemistry_5_27_html_4dd19828.gifΔG = ΔH − TΔS

Since ΔS is negative, ΔH has to be negative to make ΔG negative. Hence, adsorption is always exothermic.

*Question 5.9:

How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?

Answer:

One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Sr. No. Dispersed phase Dispersion medium Type of colloid Example
1. Solid Solid Solid Sol Gemstone
2. Solid Liquid Sol Paint
3. Solid Gas Aerosol Smoke
4. Liquid Solid Gel Cheese
5. Liquid Liquid Emulsion Milk
6. Liquid Gas Aerosol Fog
7. Gas Solid Solid foam Pumice stone
8. Gas Liquid Foam Froth

Question 5.10:

Discuss the effect of pressure and temperature on the adsorption of gases on solids.

Answer:

Effect of pressure

Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

Effect of temperature

Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.11:

What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?

Answer:

Depending upon the nature of interaction between the dispersed phase and the dispersion medium, colloidal sols are divided into two categories, namely, lyophilic (solvent attracting) and lyophobic (solvent repelling). If water is the dispersion medium, the terms used are hydrophilic and hydrophobic.

(i) Lyophilic colloids: The word ‘lyophilic’ means liquid-loving. Colloidal sols directly formed by mixing substances like gum, gelatine, starch, rubber, etc., with a suitable liquid (the dispersion medium) are called lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation), the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. Furthermore, these sols are quite stable and cannot be easily coagulated as discussed later. 

(ii) Lyophobic colloids: The word ‘lyophobic’ means liquid-hating. Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can be prepared only by special methods (as discussed later). Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence, are not stable. Further, once precipitated, they do not give back the colloidal sol by simple addition of the dispersion medium. Hence, these sols are also called irreversible sols. Lyophobic sols need stabilising agents for their preservation.

Question 5.12:

What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?

Answer:

(i)  In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii)  In macro-molecular colloids, the colloidal particles are large molecules having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, etc.

(iii)  Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called aggregated colloids.

*Question 5.13:

What are enzymes? Write in brief the mechanism of enzyme catalysis.

Answer:

Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called ‘biochemical catalysts’.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6410/NS_11-11-08_Sonali_12_Chemistry_5_27_html_m22b3450c.jpg

Mechanism of enzyme catalysis:

On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as −NH2, −COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

Hence,

Step 1: E + S → ES+

(Activated complex)

Step 2: ES+ → E + P

Question 5.14:

How are colloids classified on the basis of

(i)  Physical states of components

(ii)  Nature of dispersion medium and

(iii)  Interaction between dispersed phase and dispersion medium?

Answer:

Colloids can be classified on various bases:

(i)  On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.

(ii)  On the basis of the dispersion medium, sols can be divided as:

Dispersion medium Name of sol
Water Aquasol or hydrosol
Alcohol Alcosol
Benzene Benzosol
Gases Aerosol

(iii)  On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Question 5.15:

Explain what is observed

(i)  When a beam of light is passed through a colloidal sol.

(ii)  An electrolyte, NaCl is added to hydrated ferric oxide sol.

(iii)  Electric current is passed through a colloidal sol?

Answer:

(i)  When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

(ii)  When NaCl is added to ferric oxide sol, it dissociates to give Na+ and Cl ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl ions.

(iii)  The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.

Question 5.16:

What are emulsions? What are their different types? Give example of each type.

Answer:

The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion.

There are two types of emulsions:

(a) Oil in water type:

Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

(b) Water in oil type:

Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.

Question 5.17:

What is demulsification? Name two demulsifiers.

Answer:

The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide, etc.

Question 5.18:

Action of soap is due to emulsification and micelle formation. Comment.

Answer:

The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COONa+. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.

*Question 5.19:

Give four examples of heterogeneous catalysis.

Answer:

(i)  Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6422/NS_11-11-08_Sonali_12_Chemistry_5_27_html_365b21ab.gif

(ii)  Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6422/NS_11-11-08_Sonali_12_Chemistry_5_27_html_23418537.gif

This process is called the Haber’s process.

(iii)  Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6422/NS_11-11-08_Sonali_12_Chemistry_5_27_html_2bca8520.gif

(iv)  Hydrogenation of vegetable oils in the presence of Ni.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6422/NS_11-11-08_Sonali_12_Chemistry_5_27_html_5f263517.gif

Question 5.20:

What do you mean by activity and selectivity of catalysts?

Answer:

(a) Activity of a catalyst:

The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

(b) Selectivity of the catalyst:

The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.

(i)  https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6423/NS_11-11-08_Sonali_12_Chemistry_5_27_html_3c468527.gif

(ii)  https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6423/NS_11-11-08_Sonali_12_Chemistry_5_27_html_m1838a605.gif

(iii)  https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6423/NS_11-11-08_Sonali_12_Chemistry_5_27_html_m3b0d1324.gif

Question 5.21:

Describe some features of catalysis by zeolites.

Answer:

Zeolites are alumino-silicates that are micro-porous in nature. Zeolites have a honeycomb-like structure, which makes them shape-selective catalysts. They have an extended 3D-network of silicates in which some silicon atoms are replaced by aluminium atoms, giving them an Al−O−Si framework. The reactions taking place in zeolites are very sensitive to the pores and cavity size of the zeolites. Zeolites are commonly used in the petrochemical industry.

Question 5.22:

What is shape selective catalysis?

Answer:

A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

*Question 5.23:

Explain the following terms:

(i)  Electrophoresis  (ii)  Coagulation

(iii)  Dialysis  (iv)  Tyndall effect.

Answer:

(i) Electrophoresis:

The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation:

The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis

The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

(iv) Tyndall effect:

When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.

Question 5.24:

Give four uses of emulsions.

Answer:

Four uses of emulsions:

(i)  Cleansing action of soaps is based on the formation of emulsions.

(ii)  Digestion of fats in intestines takes place by the process of emulsification.

(iii)  Antiseptics and disinfectants when added to water form emulsions.

(iv)  The process of emulsification is used to make medicines.

Question 5.25:

What are micelles? Give an example of a micelles system.

Answer:

Micelle formation is done by substances such as soaps and detergents when dissolved in water. The molecules of such substances contain a hydrophobic and a hydrophilic part. When present in water, these substances arrange themselves in spherical structures in such a manner that their hydrophobic parts are present towards the centre, while the hydrophilic parts are pointing towards the outside (as shown in the given figure). This is known as micelle formation.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6429/NS_11-11-08_Sonali_12_Chemistry_5_27_html_1bd47e70.jpg
https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/264/6429/NS_11-11-08_Sonali_12_Chemistry_5_27_html_394bd158.jpg

Question 5.26:

Explain the terms with suitable examples:

(i) Alcosol  (ii) Aerosol  (iii) Hydrosol

Answer:

(i) Alcosol:

A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol.

For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

(ii) Aerosol:

A colloidal solution having a gas as the dispersion medium and a solid as the dispersed phase is called an aerosol.

For example: fog

(iii) Hydrosol

A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol.

For example: starch sol or gold sol

Question 5.27:

Comment on the statement that “colloid is not a substance but a state of substance”.

Answer:

Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.

 

Topics to study in  ncert chemistry class 12 chapter 5 – Surface Chemistry

Section Number

Topics

5.1

Adsorption

5.1.1

The distinction between Adsorption and Absorption

5.1.2

Mechanism of Adsorption

5.1.3

Types of Adsorption

5.1.4

Adsorption Isotherms

5.1.5

Adsorption from Solution Phase

5.1.6

Applications of Adsorption

5.2

Catalysis

5.2.1

Homogeneous and Heterogeneous Catalysis

5.2.2

Adsorption Theory of Heterogeneous Catalysis

5.2.3

Shape-Selective Catalysis by Zeolites

5.2.4

Enzyme Catalysis

5.2.5

Catalysts in Industry

5.3

Colloids

5.4

Classification of Colloids

5.4.1

Classification Based on the Physical State of Dispersed Phase and Dispersion Medium

5.4.2

Classification Based on the Nature of Interaction between Dispersed Phase and Dispersion Medium

5.4.3

Classification Based on the Type of Particles of the Dispersed Phase, Multimolecular, Macromolecular and Associated Colloids

5.4.4

Preparation of Colloids

 

5.4.5

Purification of Colloidal Solutions

5.4.6

Properties of Colloidal Solutions

5.5

Emulsions

5.6

Colloids around Us

 

Weightage of Surface Chemistry class 12 in Board exam – Term II

This chapter is going to give you 4 marks in your CBSE Term I exam. You can grab all 4 marks by practicing questions and learning concepts. By this, you will be able to solve the objective types of questions in your exam in a very short period of time.

Chapter 

Marks

Surface chemistry

4 Marks

Related Link

NCERT Solution for Class XIIth Chemistry Chapter 1 The Solid State

NCERT Solution for Class XIIth Chemistry Chapter 3 Electrochemistry

NCERT Solution for Class XIIth Chemistry Chapter 11 Alcohols, Phenols and Ethers

NCERT Solution for Class XIIth Chemistry Chapter 7 The p-Block Elements

NCERT Solution for Class XIIth Chemistry Chapter 13 Amines

NCERT Solution for Class XIIth Chemistry Chapter 9 Coordination Compounds

Conclusion

Swastik Classes’ NCERT Solutions for Class 12 Chemistry Chapter 5 – Surface Chemistry, provide a comprehensive understanding of the behavior of matter at the interface of two phases. Our solutions are designed to help students understand the concepts of adsorption, types of adsorption, and factors affecting adsorption.

Our NCERT Solutions for Class 12 Chemistry Chapter 5 also cover the applications of surface chemistry in various fields such as catalysis, colloid chemistry, and corrosion. We have provided step-by-step solutions to all the questions in the chapter, making it easy for students to understand and follow the concepts. Our solutions also include illustrations and diagrams to help students visualize the processes and remember them better.

By using our NCERT Solutions for Class 12 Chemistry Chapter 5, students can prepare for their exams, score well, and build a strong foundation in surface chemistry. Our solutions provide students with a comprehensive resource to learn and excel in the subject. Overall, our NCERT Solutions for Class 12 Chemistry Chapter 5 are an effective and comprehensive resource for students to understand and master the concepts of surface chemistry.

Why do students follow SWC’s Chemistry class 12 solutions for chapter 5?

At SWC, students directly learn from IITians and experienced faculties. We provide quality study materials to prepare and follow a LEAP model. We teach not only for exams but for lifelong applications. 

Why Swastik classes?

How to study NCERT Solution for Class 12 Chemistry Chapter 5?

Many times you sit to study the NCERT textbook in one way. It would help if you tried to grasp every concept. But this is not an efficient way to study from the NCERT textbook. With this, you will retain things only for a brief period. After that, you forgot what you had learned about that chapter. As a result, your confidence automatically decreases. This is why many students aren’t capable of understanding the NCERT textbook. To make this easy to grasp. You should follow a simple methodology while studying a particular chapter. Here are the few methods to get maximum outcome:

How to study ncert textbook

FAQs related to chemistry class 12 chapter 5

What are the important topics for surface chemistry?

Following are the important topics in surface chemistry from the CBSE board exam point of view.

  • Adsorption
  • Catalysis
  • Colloids
  • Emulsion

Students are highly advised to prepare these topics thoroughly to get grip on this chapter. So that you can perform well in exams.

 

What are the important questions in chemistry class 12 chapter 5

  • What is the coagulation process?
  • What is the difference between physisorption and chemisorption?
  • What do you mean by the Tyndall effect?
  • What is the meaning of micelles, peptization, and desorption?
  • State following terms with one example each.
    • Emulsion
    • Peptization
    • Macromolecular solution
  • Write down the difference between Adsorption and Absorption?

What are the methods of preparation of solution in surface chemistry?

Methods of preparation of solutions surface chemistry

What are the applications of adsorption?

  • Purification of water
  • Removal of moisture and humidity
  • Adsorption chromatography
  • In metallurgy
  • Removal of moisture and humidity
  • Air pollution masks

What is bulk in chemistry?

In simple, bulk is something that is available in large quantities. This definition is the same for surface chemistry whether it is used for solid-solid, liquid-gas, or liquid-liquid. As they use in large quantities (i.e bulk ) to study these different phases. 

 

swc google search e1651044504923

2021 Result Highlight of Swastik Classes

NCERT Solutions Class 12 Chemistry Chapters