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NCERT Solutions for Class 12 Chemistry Chapter 5

NCERT Solutions Class 12 Chemistry Chapter 5

Adda247 provides NCERT Solutions for Class 12 Chemistry Chapter 5. These NCERT Solutions for Class 12 Chemistry Chapter 5 will not only help students to boost their board exams and score brilliant marks but for competitive exams. The NCERT Solutions for Class 12 Chemistry Chapter 5 are according to the NCERT guidelines.

Complete 16 chapters wise solutions were provided for the benefit of students.12th class sets the base for higher education for every student. This makes it the most crucial class for any student who is aiming for his/her dream of quality education. Scoring good marks in the 12th class is equivalent to quality high education. So, it becomes extremely important for students to give a boost to their class 12 Chemistry preparation with Adda247 NCERT solutions.

Class 12 is very crucial in a student’s life because the marks score in class 12 plays a vital role in getting admission to the college of your own choice. Moreover, the marks scored in class 12 are also important to get admission to many government exams.

The students can access the solutions anywhere while browsing the web easily. The NCERT Solutions for Class 12 Chemistry Chapter 5 are very precise and accurate.

NCERT Solutions for Class 12 Chemistry Chapter 5 includes a number of topics that offer students an in-depth knowledge of different aspects of chemistry. To distinguish the equations and chemical formulas, solving the NCERT chemistry class 12 PDF can be handy.


NCERT Solutions For Class 12 Chemistry Chapter 5: Surface Chemistry  

The fifth chapter in chemistry class 12 gives a brief idea about the chemical reaction related to solid-liquid, solid-gas, and liquid gas.

Surface chemistry deals with the phenomenon that occur at the surface or interface. The interface or surface is represented by separating the bulk phase by a hyphen or a slash. For example, the interface between a solid and a gas may be represented by solid-gas or solid/gas.

NCERT Solutions Class 12 Chemistry Chapter 5 PDF

Surface chemistry plays an important role in everyday life, as the basis for many phenomenon as well as technological applications. Common examples range from soap bubbles, foam, and raindrops to cosmetics, paint, adhesives, and pharmaceuticals.

Surface chemistry is important in many critical chemical processes, such as enzymatic reactions at biological interfaces found in cell walls and membranes, in electronics at the surfaces and interfaces of microchips used in computers, and the heterogeneous catalysts found in the catalytic converter used for cleaning.


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NCERT Solutions for Class 12 Chemistry Chapter 5: Surface Chemistry Key Features

  • The NCERT Solution provide clear and precise answer.
  • The columns are used wherever necessary.
  • We follow the NCERT guidelines.
  • Illustration through diagrams.
  • Adequate examples for better explanation.


Class 12 Chemistry Chapter 5 NCERT Solutions: Surface Chemistry Important Questions

Que:1  Distinguish b/w the meaning of the terms adsorption and absorption. Give one example of each.


The accumulation of molecular species at the surface rather than in bulk of solid or liquid is termed as adsorption. It is a surface phenomenon. Solids, particularly in finely divided state, have large surface area and therefore, alumina gel, silica gel, clay, metals in finely divided state etc. act as good adsorbent. For example, when we dip a chalk stick into and ink solution, only its surface becomes coloured, if we break the chalk stick, it will be found to be white from inside.

In absorption, the substance is uniformly distributed throughout the bulk of the solid. Both adsorption and absorption can take place simultaneously

Que:2 Give a reason why finally divided substances more effective as an adsorbent.


Adsorption  surface area.                              

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


Que:3 What is the difference b/w physisorption and chemisorption.


             Physisorption               Chemisorption
1.  It arises because of van der waal’s                force. 1. It is caused by chemical bond formation.
2. It is reversible in nature. 2. It is irreversible.
3. It is not specific in nature. 3. It is highly specific in nature.
4. Enthalpy of adsorption is low. 4. Enthalpy of adsorption is high.
5. It depends on the nature of gas. More easily liquefiable gases are adsorbed readily. 5. It also depends on the nature of gas. Gases which can react with the adsorbent show chemisorption.
6. No appreciable activation energy is needed. 6. High activation energy is sometimes needed.


Que:4 What are the factors which influence the adsorption of a gas on a solid?


Following are the 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 H2O2 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 reversible process and is accompanied by decrease in pressure. Therefore, adsorption increase 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.


Que:5 What do you understand by activation of adsorbent? How is it achieved?


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

By increasing the surface area. This can be done by breaking the substance into small pieces or powdering it.

Some specific treatment also to the activation of an adsorbent. For eg., wood charcoal is activated by heating it b/w 650K and 1330K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.


Que:6 Why is adsorption always exothermic?


i.) Adsorption leads to a decrease in 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.)H of adsorption is always negative. When a gas is adsorbed on a solid     surface, its movement is restricted leading to a decrease in entropy of the gas i.e. S is negative. Now for a process to be spontaneous, G should be negative.

: G = H – TS

Since S is negative H has to be negative to make G negative. Hence, adsorption is always exothermic.


Que:7 How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?


Depending upon whether the dispersed phase and the  dispersion medium are solids, liquids or gases, eight types of colloidal system are possible.

Dispersed phase Dispersion medium


Type of Colloid






Solid Solid Solid sol Gem stones
Solid Liquid Sol Paints, cell fluids
Solid Gas Aerosol Smoke, dust
Liquid Solid Gel Cheese, butter
Liquid Liquid Emulsion Milk, hair
Liquid Gas Aerosol Fog, mist





Solid sol


Foam rubber

Froth, soap lather


Que:8 Discuss the effect of pressure and temperature on the adsorption of gases on solids.


Effect of pressure: Adsorption is reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increase 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 decrease with an increase in temperature


Que:9 What is the difference b/w multi – molecular and macro – molecular colloids? Give one example of each. How are associated colloids different from these two types of colloids


  • In multi-molecular collides, the colloidal particles are and 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. Example of such colloids includes gold sol and sulphur sol.


  • 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 e.g.: starch, nylon, etc.


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


Que:10 What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?


i.) Lyophilic sols: Colloid sols that are formed by mixing substances such as gum, starch, etc. with a suitable liquid are called lyophilic sols. These sols are reversible in nature i.e. if two constituents of the sol are separated by any means, than the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture.


ii.)Lyophobic sols: When substance such as metals and their sulphides etc. mixed with the dispersion medium they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature. For example: Sols of metals. Now the stability of hydrophilic sols depends on two things the pressure of a charge and the salvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of pressure of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed, than the particles present in them come c loser and form aggregates, leading to precipitation.

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Ncert Solutions for Class 12 Chemistry Chapter 5: FAQs

  1. What are the important questions from Class 12 Surface Chemistry?

Ans. A few important questions from Class 12 Surface Chemistry are given as follows:

  1. Write any two characteristics of chemisorption.
  2. Why is it necessary to remove CO when Haber’s process obtains ammonia?
  3. What is the role of desorption in the process of catalysis?
  4. Why is it essential to wash the precipitate with water before estimating it quantitatively?
  5. Give a reason why a finely divided substance is a good as absorbent?
  6. What are the factors which influence the adsorption of a gas on a solid?

All the questions are essential for the students to be studied in detail to answer questions from both the short and very short questions categories.

  1. How helpful is the NCERT Solution for chapter 5 Surface Chemistry?

Ans. The NCERT Solution deals with all difficult topics in Chapter 5. The solution will help students understand the definitions and classifications very easily. This also enables the students to learn important terms like emulsification, micelles dialysis, coagulation, adsorption, and absorption. Topics like why the powdered substance is better adsorbent than crystalline ones are also delved in detail through elaborate illustrations and diagrams.

The NCERT Solutions also give answers to situational questions where students are required to think analytically. The questions are all answered by expert Chemistry teachers who provide greater insights into the critical concepts of the Chapter.

  1. How to access NCERT PDF Solutions?

Ans. Student can download the NCERT Solutions for class 12 chapter 5 PDF from e-learning websites online for free. Students need to register on the official websites to access the files. Once registered, the students will be able to download the PDF version of the NCERT solutions for all the subjects chapter wise. The PDF of solutions contain detailed explanations of important topics with illustrations and diagrams. The solutions are updated on a regular basis as per the CBSE syllabus to help students score mare marks in the board exams. By regular practice, students will also be able to improve their time management skills which are important from the exams point of view.

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