NCERT Exemplar Class 12 Chemistry Solutions Chapter 5 Surface Chemistry

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5 Surface Chemistry

Edited By Shivani Poonia | Updated on Apr 05, 2025 12:58 PM IST | #CBSE Class 12th

Surface chemistry deals with the intriguing world of reactions occurring at the interface of different phases. It talks about the properties, mechanisms, and applications of surface phenomena that are crucial in industries, catalysis and biological systems as well. Adsorption, catalysis, colloids, emulsions, and surfactants are the important topics that are discussed in this chapter. Numerous industrial operations together with everyday life implementations, use these principles. Industrial reactions benefit from catalysts that enhance their performance, as in the Haber and Contact processes. Water purification as well as gas masks and drug delivery systems, employ adsorption for their operation. The development of nanotechnology alongside self-cleaning surfaces emerges from surface chemistry principles.

Textbook of NCERT presents these concepts in a structured manner by introducing the NCERT Exemplar Class 12 Chemistry Solutions that gives well-explained answers, numerical problems, and conceptual questions for better understanding and application-based learning. Mastering surface chemistry is a must for students preparing for board exams and competitive exams.

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: MCQ (Type 1)
  2. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: MCQ (Type 2)
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Short Answer Type
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Matching Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Assertion and Reason Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Long Answer Type
  7. Class 12 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS) Questions
  8. Approaches to Solve Questions of Chapter 5 Surface Chemistry
  9. Following are the major topics in Class 12 Chemistry NCERT Exemplar Chapter 5
  10. NCERT Exemplar Class 12 Chemistry Chapter 5 Surface Chemistry - Learning Outcome
  11. NCERT Exemplar Class 12 Chemistry Solutions Chapter-wise
  12. Check NCERT Solutions for Class 12 Chemistry
NCERT Exemplar Class 12 Chemistry Solutions Chapter 5 Surface Chemistry
NCERT Exemplar Class 12 Chemistry Solutions Chapter 5 Surface Chemistry
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NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: MCQ (Type 1)

Question:1

Which of the following process does not occur at the interface of phases?
(i) crystallisation
(ii) heterogeneous catalysis
(iii) homogeneous catalysis
(iv) corrosion
Answer:

The answer is the option (iii). There is no homogeneous catalysis at the interface of phases since in the case of homogenous catalysis reactant and catalyst have same phase and a uniform distribution.

Question:2

At the equilibrium position in the process of adsorption ___________.
$(i)\; \Delta H >0$
$(ii)\; \Delta H =T\Delta S$
$(iii)\; \Delta H >T\Delta S$
$(iv)\; \Delta H <T\Delta S$
Answer:

The answer is the option (ii). $\text {At equilibrium position during adsorption, }\Delta G=\Delta H-T\Delta S=0$ So that $\Delta H=T\Delta S$

Question:3

Which of the following interface cannot be obtained?
(i) liquid-liquid
(ii) solid-liquid
(iii) liquid-gas
(iv) gas-gas
Answer:

The answer is the option (iv). As gases are completely miscible, they do not form a gas-gas interface. Air, for instance, contains Nitrogen, Oxygen, Carbon dioxide, etc.

Question:4

The term ‘sorption’ stands for ____________.
(i) absorption
(ii) adsorption
(iii) both absorption and adsorption
(iv) desorption
Answer:

The answer is the option (iii). Sorption is the simultaneous occurrence of adsorption and absorption.

Question:5

Extent of physisorption of a gas increases with ___________.
(i) increase in temperature.
(ii) the decrease in temperature.
(iii) the decrease in surface area of the adsorbent.
(iv) the decrease in strength of van der Waals forces.
Answer:

The answer is the option (ii).
Physisorption of a gas increases with decrease in temperature. This is because, weak van der Waals forces hold the physiosorption particles to the surface making them difficult to exist at higher temperature.

Question:6

The extent of adsorption of adsorbate from the solution phase increases with ________.
(i) increase in the amount of adsorbate in solution.
(ii) decrease in surface area of the adsorbent.
(iii) increase in temperature of the solution.
(iv) decrease in the amount of adsorbate in solution
Answer:

The answer is the option (i).
The molecular species that get adsorbed on the surface is known as adsorbate. With an increase in the amount of adsorbate, both the interaction of adsorbent and the extent of adsorption increase.

Question:7

Which one of the following is not applicable to the phenomenon of adsorption?
$(i)\; \Delta H >0$
$(ii)\; \Delta G <0$
$(iii)\; \Delta S <0$
$(iv)\; \Delta H <0$
Answer:

The answer is the option (i).
Since, adsorption is an exothermic process $\Delta H\; >\; 0$

Question:8

Which of the following is not a favourable condition for physical adsorption?
(i) high pressure
(ii) negative $\Delta \text {H}$
(iii) the higher critical temperature of adsorbate
(iv) high temperature
Answer:

The answer is the option (iv).
Adsorption is an exothermic process. The rate of adsorption decreases with an increase in temperature.

Question:9

Physical adsorption of a gaseous species may change to chemical adsorption with ______________.
(i) decrease in temperature
(ii) increase in temperature
(iii) increase in surface area of the adsorbent
(iv) the decrease in surface area of the adsorbent
Answer:

The answer is the option (ii).
If temperature increases, chemical bonds start forming between the adsorbate and adsorbent. It is so because the energy for activation is reached on increasing the temperature. Hence, physical adsorption of gaseous species may change to chemical adsorption.

Question:10

In physisorption, adsorbent does not show specificity for any particular gas because ______________.
(i) involved van der Waals forces are universal.
(ii) gases involved behave like ideal gases.
(iii) enthalpy of adsorption is low.
(iv) it is a reversible process.
Answer:

The answer is the option (i).
Physisorption is a byproduct of Van der Waal’s forces between adsorbate and adsorbent, which are universal and hence, physisorption is independent of the gas chosen.

Question:11

Which of the following is an example of absorption?
(i) Water on silica gel
(ii) Water on calcium chloride
(iii) Hydrogen on finely divided nickel
(iv) Oxygen on the metal surface

Answer:

The answer is the option (ii).
Calcium chloride absorbs water. Water on silica gel, hydrogen on finely divided nickel and oxygen on metal surface are examples of absorption

Question:12

Based on data given below predict which of the following gases shows the least adsorption on a definite amount of charcoal?

Gas
Critical temp./K

$\text {CO}_{2}$
304

$\text {SO}_{2}$
630

$\text {CH}_{4}$
190

$\text {H}_{2}$
33

$(i)\; \text {CO}_{2}$
$(ii)\; \text {SO}_{2}$
$(iii)\; \text {CH}_{4}$
$(iv)\; \text {H}_{2}$

Answer:

The answer is the option (iv). The extent of adsorption is higher for gases with lower critical temperature. In the above case, hydrogen has the least value of critical temperature and thus, shows the least susceptibility towards adsorption

Question:14

At high concentration of soap in water, soap behaves as ____________.
(i) molecular colloid
(ii) associated colloid
(iii) macromolecular colloid
(iv) lyophilic colloid
Answer:

The answer is the option (ii).
Soap behaves as a strong electrolyte in lower concentrations. However, as the concentration increases, it exhibits colloidal behavior due to formation of micelles (aggregates). This behavior is shown by a class of colloids known as associated colloids.

Question:15

Which of the following will show the Tyndall effect?
(i) Aqueous solution of soap below critical micelle concentration.
(ii) Aqueous solution of soap above critical micelle concentration.
(iii) Aqueous solution of sodium chloride.
(iv) Aqueous solution of sugar.
Answer:

The answer is the option (ii).
Over the critical micelle concentration, an aqueous solution of soap forms a colloidal solution. When light is passed through a colloid, it scatters due to its interaction with the particles. This effect is known as the Tyndall effect.

Question:16

The method by which lyophobic sol can be protected ?
(i) By the addition of oppositely charged sol.
(ii) By the addition of an electrolyte.
(iii) By the addition of lyophilic sol.
(iv) By boiling.

Answer:

The answer is the option (iii).
Lyophobic sols, on addition of small number of electrolytes, are readily precipitated. Adding lyophilic sol makes them stable.

Question:17

Freshly prepared precipitate sometimes gets converted to colloidal solution by ___________.
(i) coagulation
(ii) electrolysis
(iii) diffusion
(iv) peptisation
Answer:

The answer is the option (iv).
Peptisation is the process of converting freshly prepared precipitate into colloidal solution.

Question:18

Which of the following electrolytes will have maximum coagulating value for ${AgI}/{Ag^{+}}$ sol?
$(i) \; \text {Na}_{2}\text {S}$
$(ii) \; \text {Na}_{3}\text {PO}_{4}$
$(iii) \; \text {Na}_{2}\text {SO}_{4}$
$(iv) \; \text {NaCl}$
Answer:

The answer is the option (ii).
According to Hardy-Schulze law, the greater the charge on anion or the valency of coagulating ion, the greater will be its coagulating power.

Electrolytes

Anionic part

Charge on anion

$\text {Na}_{2}\text {S}$

$\text {S}^{2-}$

2

$\text {Na}_{3}\text {PO}_{4}$

$\text {PO}_{4}^{3-}$

3

$\text {Na}_{2}\text {SO}_{4}$

$\text {SO}_{4}^{2-}$

2

$\text {NaCl}$

$\text {Cl}^{-}$

1

Here, $\text {PO}_{4}^{3-}$ have highest charge. Hence, $\text {PO}_{4}^{3-}$ have highest coagulating power.

Question:19

A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is classified as ____________.
(i) solid sol
(ii) gel
(iii) emulsion
(iv) sol

Answer:

The answer is the option (iv).
A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is called sol.

Question:20

The values of colligative properties of the colloidal solution are of small order in comparison to those shown by true solutions of same concentration because of colloidal particles __________________.
(i) exhibit an enormous surface area.
(ii) remain suspended in the dispersion medium.
(iii) form lyophilic colloids.
(iv) are comparatively less in number
Answer:

The answer is the option (iv).
Colloidal particles are less in number due to their large size. Hence option d is correct.

Question:21

Arrange the following diagrams in the correct sequence of steps involved in the mechanism of catalysis, in accordance with modern adsorption theory.


$(i)\; a \rightarrow b \rightarrow c \rightarrow d \rightarrow e$
$(ii)\; a \rightarrow c \rightarrow b \rightarrow d \rightarrow e$
$(iii)\; a \rightarrow c \rightarrow b \rightarrow d \rightarrow e$
$(iv)\; a \rightarrow b \rightarrow c \rightarrow e \rightarrow d$

Answer:

The answer is the option $(ii)\; a \rightarrow c \rightarrow b \rightarrow d \rightarrow e$
i - denotes the surface has adsorbed A and B
$iii\rightarrow ii$ – denotes that A and B interact with each other to form the intermediate
$ii\rightarrow iv$ – denotes the desorption of the new molecule (A-B)
$iv\rightarrow v$ – denotes the completion of the desorption process

Question:22

Which of the following process is responsible for the formation of the delta at a a place where rivers meet the sea?
(i) Emulsification
(ii) Colloid formation
(iii) Coagulation
(iv) Peptisation

Answer:

The answer is the option (iii).
While river water is cleaner, sea water is salty. It is because of this that the place where river meets the sea colloidal particles settle down to form delta.

Question:23

Which of the following curves is in accordance with Freundlich adsorption isotherm?

Answer:

The answer is the option (iii).
The freundlich adsorption isotherm graph gives:
$\frac{\text {x}}{\text {m}}=\text {kp}^{\frac{1}{\text {n}}}$
Or $\text {log}\frac{\text {x}}{\text {m}}=\text {log k}+{\frac{1}{\text {n}}}\text {log p}$

Question:24

Which of the following process is not responsible for the presence of electric charge on the sol particles?
(i) Electron capture by sol particles.
(ii) Adsorption of ionic species from solution.
(iii) Formation of Helmholtz electrical double layer.
(iv) Absorption of ionic species from solution.

Answer:

The answer is the option (iv).
Charge on sol particles is mostly due to the preferential adsorption of either positive or negative ions on their surface. Although there is no absorption of ionic species. Other processes responsible for presence of electric charge on sol particles include electron capture by sol particles and formation of Helmholtz electrical double layer.

Question:25

Which of the following phenomenon applies to the process shown in Fig. 5.1?

(i) Absorption
(ii) Adsorption
(iii) Coagulation
(iv) Emulsification

Answer:

The answer is the option (ii).
Sugar solution which comes out from the column is colourless because animal charcoal adsorbs the colour.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: MCQ (Type 2)

Question:26

Which of the following option are correct?
(i) Micelle formation by soap in aqueous solution is possible at all temperatures.
(ii) Micelle formation by soap in aqueous solution occurs above a particular concentration.
(iii) On dilution of soap, solution micelles may revert to individual ions.
(iv) Soap solution behaves like a normal strong electrolyte at all concentrations.

Answer:

The answer is the option (ii, iii) Micelle formation is possible only above critical micelle concentration and on dilution of soap solution the micelles may form electrolytes.

Question:27

Which of the following statements are correct about solid catalyst?
(i) Same reactants may give different product by using different catalysts.
(ii) Catalyst does not change $\Delta \text {H}$ of reaction.
(iii) Catalyst is required in large quantities to catalyse reactions.
(iv) The catalytic activity of a solid catalyst does not depend upon the strength of chemisorption.

Answer:

The answer is the option (i, ii)
Catalysts are highly selective in nature. They may act as a catalyst for a particular reaction but might not have any effect on the other. In addition, using different catalysts for same reaction may yield different products. Catalyst does not change the enthalpy of reactions. Enthalpy of a reaction is defined as the difference between the enthalpy of reactants and products. Therefore, catalyst has no role to play in the enthalpy of a reaction.

Question:28

Freundlich adsorption isotherm is given by the expression $\frac{\text {x}}{\text {m}}=\text {kp}^{\frac{\text {1}}{\text {n}}}$ which of the following conclusions can be drawn from this expression
(i) When ${\frac{\text {1}}{\text {n}}}=0,$, the adsorption is independent of pressure.
(ii) When ${\frac{\text {1}}{\text {n}}}=0,$ , the adsorption is directly proportional to pressure.
(iii)When $\text {n}=0, \frac{\text {x}}{\text {m}}$ vs p graph is a line parallel to x-axis.
(iv) When $\text {n}=0,$,a plot of $\frac{\text {x}}{\text {m}}$ vs p is a curve.

Answer:

The answer is the option (i, iii)

Question:29

$\text {H}_{2}$ gas is adsorbed on activated charcoal to a very little extent in comparison to easily liquefiable gases due to ____________.
(i) very strong van der Waal’s interaction.
(ii) very weak van der Waals forces.
(iii) very low critical temperature.
(iv) very high critical temperature.
Answer:

The answer is the option (ii, iii). Due to very low critical temperature, only limited amount of hydrogen is adsorbed.

Question:30

Which of the following statements are correct?
(i) Mixing two oppositely charged sols neutralises their charges and stabilises the colloid.
(ii) Presence of equal and similar charges on colloidal particles provides stability to the colloids.
(iii) Any amount of dispersed liquid can be added to emulsion without destabilising it.
(iv) Brownian movement stabilises sols.

Answer:

The answer is the option (ii, iv). Colloidal particles have equal and similar charges because of which they repel each other. Brownian movement does not allow the colloidal sols to settle down. Hence, it helps in providing stability.

Question:31

An emulsion cannot be broken by __________ and ___________.
(i) heating
(ii) adding more amount of dispersion medium
(iii) freezing
(iv) adding emulsifying agent

Answer:

The answer is the option (ii, iv) Adding more amount of dispersion medium or emulsifying agent can’t break an emulsion because addition of dispersion medium dilutes the emulsion and emulsifying agent stabilizes it. It can be broken by heating, freezing and centrifuging.

Question:32

Which of the following substances will precipitate the negatively charged emulsions?
(i) $\text {KCl}$
(ii) glucose
(iii) urea
(iv) $\text {NaCl}$
Answer:

The answer is the option (i, iv) Negatively charged emulsions will precipitate because of the dissociation products $\text {K}^{+}$ and $\text {Na}^{+}$ from $\text {KCl}$ and $\text {NaCl}.$

Question:33

Which of the following colloids cannot be coagulated easily?
(i) Lyophobic colloids.
(ii) Irreversible colloids.
(iii) Reversible colloids.
(iv) Lyophilic colloids.

Answer:

The answer is the option (iii, iv) Coagulation of lyophilic colloids, also known as reversible colloids, isn’t easy as they are already stable.

Question:34

What happens when a lyophilic sol is added to a lyophobic sol?
(i) Lyophobic sol is protected.
(ii) Lyophilic sol is protected.
(iii) Film of lyophilic sol is formed over lyophobic sol.
(iv) Film of lyophobic sol is formed over lyophilic sol.
Answer:

The answer is the option (i, iii) Lyophilic sols are more stable than lyophobic sols. On addition of lyophilic sols to lyophobic sols, lyophobic sols are protected due to formation of film of lyophilic sol.

Question:35

Which phenomenon occurs when an electric field is applied to a colloidal solution and electrophoresis is prevented?
(i) Reverse osmosis takes place.
(ii) Electroosmosis takes place.
(iii) Dispersion medium begins to move.
(iv) Dispersion medium becomes stationary.
Answer:

The answer is the option (ii, iii) When an electric field is applied to a colloidal solution and electrophoresis is prevented electroosmosis takes place.

Question:36

In a reaction, catalyst changes ____________.
(i) physically
(ii) qualitatively
(iii) chemically
(iv) quantitatively

Answer:

The answer is the option (i, ii). Catalyst does not participate in the reaction and thus doesn’t change chemically or quantitatively.

Question:37

Which of the following phenomenon occurs when a chalk stick is dipped in ink?
(i) adsorption of coloured substance
(ii) adsorption of solvent
(iii) absorption and adsorption both of solvent
(iv) absorption of solvent
Answer:

The answer is the option (i, iv) Chalk adsorbs the coloured substance, but absorbs the solvent.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Short Answer Type

Question:38

Why is it important to have a clean surface in surface studies?

Answer:

Adsorption is a surface phenomenon that occurs only at the surface of an adsorbent. An unclean surface will reduce the surface area available to carry out adsorption.

Question:39

Why is chemisorption referred to as activated adsorption?

Answer:

Chemisorption involves formation of bonds between the solid surface and gaseous atoms/molecules. It requires activation energy to create the bonds and this is the reason it is known as activated adsorption.

Question:40

What type of solutions are formed on dissolving different concentrations of soap in the water?
Answer:

The solution of soap in water at low concentration is an electrolytic solution. However, on increasing the concentration of soap, it reaches the critical micelle concentration. If the concentration of soap exceeds further, it becomes a colloidal solution.

Question:41

What happens when gelatin is mixed with gold sol?

Answer:

Gold sol is lyophobic sol or solvent repelling sol. It is unstable in nature. Gelatin stabilizes the colloidal solution of gold by acting as a protective colloid.

Question:42

How does it become possible to cause artificial rain by spraying silver iodide on the clouds?
Answer:

Clouds carry a charge and are colloidal. Silver Iodide interacts with the colloidal particles, neutralizes the particles, and coagulates the clouds, resulting in rain.

Question:43

Gelatin which is a peptide is added in ice creams. What can be its role?
Answer:

Gelatin acts as an emulsifying agent in Ice cream (an emulsion) and stabilizes it. It is used to help the ice cream get a smooth texture by maintaining its consistency.

Question:44

What is collodion?
Answer:

Collodion is a solution of 4% cellulose nitrate in alcohol-ether mixture and is used to filter colloidal solution.

Question:45

Why do we add alum to purify water?

Answer:

Alum is used to coagulate negatively charged colloidal particles present in river water, which leads to settling of the suspended material.

Question:46

What happens when the electric field is applied to a colloidal solution?
Answer:

Colloidal particles carry a charge and on application of electric field, they move towards oppositely charged electrodes. This phenomenon of clearing colloidal particles is known as electrophoresis and is of two types cataphoresis and anaphoresis.

Question:47

What causes Brownian motion in colloidal dispersion?
Answer:

Colloidal particles move in a continuous zigzag movement in a colloidal sol. This random movement of the colloidal particles is known as Brownian motion. Colloidal particles would normally settle down, but due to collisions between the colloidal particles (which results in Brownian motion) the particles do not settle and stabilizes the colloidal solution.

Question:48

A colloid is formed by adding $\text {FeCl}_{3}$ in excess of hot water. What will happen if excess sodium chloride is added to this colloid?
Answer:

On adding excess sodium chloride to a colloid formed by adding $\text {FeCl}_{3}$ in excess of hot water, coagulation occurs. It happens because the chloride ions from sodium chloride neutralizes the positively charged hydrated ferric oxide solution.

Question:49

How do emulsifying agents stabilise the emulsion?
Answer:

Between the dispersion medium and the suspended particles, an interfacial layer is formed by the emulsifying agent. Emulsifiers coat droplets within an emulsion to prevent them from coming together and thus stabilizing the emulsion.

Question:50

Why are some medicines more effective in the colloidal form?
Answer:

Due to the large surface area of the colloids and easy assimilation and absorption, effectiveness of some medicines increases in colloidal state.

Question:51

Why does leather get hardened after tanning?
Answer:

Animal skin carries a positive charge and is colloidal in nature. Upon interaction with Tannin, which is a negatively charged colloidal solution, they coagulate and this phenomenon results in the hardening of leather.

Question:52

How does the precipitation of colloidal smoke take place in Cottrell precipitator?
Answer:

Smoke is the colloidal solution of the solid particles such as carbon, arsenic, dust particles etc. When smoke particles (which are charged) pass through the chamber in Cottrell Capacitor, which have oppositely charged plates, they are attracted to the plates and precipitate upon losing charge (due to contact with the plate).

Question:53

How will you distinguish between the dispersed phase and dispersion medium in an emulsion?
Answer:

Adding dispersion medium to emulsions will dilute it. Excessive amount of dispersed phase, on addition to the emulsion, forms a separate layer. This is how we distinguish between dispersed phase and dispersion medium.

Question:54

Based on Hardy-Schulze rule explain why the coagulating power of phosphate is higher than chloride.
Answer:


Coagulating value is defined as the least quantity of electrolyte which can lead to the precipitation of a sol. The charge on phosphate is higher than the charge on chloride. Coagulating power of an electrolyte ‘A’ is higher than ‘B’ if –

  • Smaller amount of ‘A’ can cause precipitation

  • ‘A’ has a higher charge

Question:55

Why does the bleeding stop by rubbing moist alum?
Answer:

Blood is a colloidal solution when we rub moist alum on that part it causes coagulation of blood leading to the formation of blood clots which stops bleeding.

Question:56

Why is $\text {Fe (OH)}_{3}$ colloid positively charged, when prepared by adding $\text {FeCl}_{3}$ to hot water?
Answer:

Due to adsorption of Ferric $\text {Fe}^{3+}$ ions by the hydrated ferric oxide solution, the colloidal solution becomes positively charged.

Question:57

Why do physisorption and chemisorption behave differently with rising in temperature?
Answer:

While physisorption occurs because of van der waal’s forces, which weaken with increasing temperature, chemisorption occurs by forming bonds. On increasing temperature, the activation energy required to form the bonds is reached more easily. Hence, physisorption weakens with rising temperature, but chemisorption strengthens.

Question:58

What happens when dialysis is prolonged?
Answer:

When dialysis is prolonged, electrolyte traces responsible for stabilising the colloids, are completely removed. This leads to the the colloid becoming unstable and hence, coagulation occurs.

Question:59

Why does the white precipitate of silver halide become coloured in the presence of dye eosin?
Answer:

Pigments of eosin dye are adsorbed by the white precipitate of silver chloride giving a colour to the precipitate.

Question:60

What is the role of activated charcoal in a gas mask used in coal mines?
Answer:

Activated charcoal means that it has been heated to increase its absorbent activity. In gas mask, charcoal adsorbs the poisonous gases like $\text {SO}_{2}, \text {NO}_{2},\text {CH}_{4},\text {CO}_{2}$, etc. present in atmosphere due to heavy traffic and provides fresh oxygen for inhaling.

Question:61

How does a delta form at the meeting place of sea and river water?
Answer:

While river water is cleaner, sea water is salty. It is because of this that the place where river meets the sea colloidal particles settle down to form delta.

Question:62

Give an example where physisorption changes to chemisorption with rise in temperature.Explain the reason for change.
Answer:

If temperature increases, chemical bonds start forming between the adsorbate and adsorbent. It is so because the energy for activation is reached on increasing the temperature. Hence, physisorption of gaseous species may change to chemisorption. At lower temperatures, hydrogen is adsorbed by weaker van der Waals forces to finely divided nickel. However, when the temperature increases, bonds start to form between Nickel and hydrogen.

Question:63

Why is desorption important for a substance to act as a good catalyst?
Answer:

Adsorption is directly proportional to the free surface are available. If reactants are not removed from the surface properly, adsorption of the next batch of reactants will be impacted negatively and will hamper the reaction.

Question:64

What is the role of diffusion in heterogeneous catalysis?
Answer:

Heterogenous catalyst is used to increase the rate of reaction in which catalyst is not in phase with reactants and products. Gaseous molecules get adsorbed on the surface of the catalyst. After the chemical reaction has taken place, products are desorped from the surface. New molecules can then occupy the freed-up space.

Question:65

How does a solid catalyst enhance the rate of combination of gaseous molecules?
Answer:

The rate of reaction increases proportionally with an increase in concentration of reactants. The concentration of gaseous molecules increases at the surface of the catalyst on getting adsorbed. The rate of reaction increases by adsorption of different molecules side by side thus facilitating the chemical reaction. In addition, adsorption process is exothermic. Therefore, it releases energy which helps in further increasing the rate of reaction.

Question:66

Do the vital functions of the body such as digestion get affected during fever? Explain your answer.
Answer:

Enzyme reactions require an optimal temperature range. An increase or decrease in the temperature from this value will reduce the enzyme activity. This optimum temperature ranges between 298K and 310K. A person suffering from fever has their body temperature in excess of 310K, which impacts the enzymatic reactions negatively.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Matching Type

Question:67

Method of formation of solution is given in Column I. Match it with the type of solution given in Column II.

Column I

Column II

(i)

Sulphur vapour passed through cold water

(a)

Normal electrolyte solution

(ii)

Soap mixed with water above critical micelle concentration

(b)

Molecular colloids

(iii)

White of egg whipped with water

(c)

Associated colloid

(iv)

Soap mixed with water below critical micelle concentration

(d)

Macro-molecular colloids

Answer:

(i - b), (ii-c), (iii-d), (iv-a)
(i) Molecular colloids are formed when sulphur vapour is passed through cold water.
(ii) Associated colloids are formed when soap is mixed with water and the concentration exceeds critical micelle concentration.
(iii)Macro-mollecular colloids are formed when the white of egg is whipped with water.
(iv) Normal electrolyte solution is formed when soap is mixed with water and the concentration is below the critical micelle concentration.

Question:68

Match the statement given in Column I with the phenomenon given in Column II.

Column I

Column II

(i)

Dispersion medium moves in an electric field

(a)

Osmosis

(ii)

Solvent molecules pass through semipermeable membrane towards solvent side

(b)

Electrophoresis

(iii)

Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes

(c)

Electroosmosis

(iv)

Solvent molecules pass through semipermeable membranes towards the solution side

(d)

Reverse osmosis

Answer:

(i→ c), (ii→ d), (iii→ b), (iv→ a)
(i) When dispersion medium moves in an electric field, it is known as electroosmosis.
(ii) When solvent molecules pass through semipermeable membrane towards
solvent side, it is known as reverse osmosis.
(iii) Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charge electrodes is known as electrophoresis.
(iv) Solvent molecules pass through semipermeable membranes towards solution side is known as osmosis.

Question:69

Match the items given in Column I and Column II.

Column I

Column II

(i)

Protective colloid

(a)

$\text {FeCl}_{3}+\text {NaOH}$

(ii)

Liquid-liquid colloid

(b)

Lyophilic colloids

(iii)

Positively charged colloid

(c)

Emulsion

(iv)

Negatively charged colloid

(d)

$\text {FeCl}_{3}+\text {hot water}$

Answer:

(i - b), (ii - c), (iii - d), (iv- a)
(i) Lyophilic colloids make a protective layer around the lyophobic sol stabilizing them. Lyophobic sols can also be protected by simply adding small doses of electrolyte.
(ii) Partially miscible or immiscible liquids can form a liquid liquid colloid (emulsion)
(iii) Positively charged colloid is formed on addition of $\text {FeCl}_{3}$ to hot water.
(iv) Negatively charged colloid is formed on addition of $\text {FeCl}_{3}$ to $\text {NaOH}$

Question:70

Match the types of colloidal systems given in Column I with the name given in Column II.

Column I

Column II

(i)

Solid in liquid

(a)

Foam

(ii)

Liquid in solid

(b)

Sol

(iii)

Liquid in liquid

(c)

Gel

(iv)

Gas is liquid

(d)

Emulsion

Answer:

(i -b), (ii —c), (iii — d), (iv — a)
Solid in Liquid colloid is a Sol; Liquid in Solid colloid is a gel; Liquid in Liquid colloid is an emulsion; Gas in Liquid colloid is foam

Question:71

Match the items of Column I and Column II.

Column I

Column II

(i)

Dialysis

(a)

Cleansing action of soap

(ii)

Peptisation

(b)

Coagulation

(iii)

Emulsification

(c)

colloidal sol formation

(iv)

Electrophoresis

(d)

Purification

Answer:

(i - d), (ii - c), (iii - a), (iv - b)
(i) Dialysis can purify colloids as a semi-permeable membrane is used to remove charged particles from the mix.
(ii) Peptisation is a process of converting a precipitate into colloidal particles by adding suitable electrolyte.
(iii) Emulsification helps to remove dirt (oily or greasy) from cloth
(iv) Electrophoresis is a process which separates charged particles in a fluid using a field of electrical charge. This process can be used to coagulate colloids.

Question:72

Match the items of Column I and Column II.

Column I

Column II

(i)

Butter

(a)

Dispersion of liquid in liquid

(ii)

Pumice stone

(b)

Dispersion of solid in liquid

(iii)

Milk

(c)

Dispersion of gas in solid

(iv)

Paints

(d)

Dispersion of liquid in solid

Answer:

(i — d), (ii — c), (iii — a), (iv — b)
(i) Butter – dispersion of liquid in solid.
(ii) Pumice stone – dispersion of gas in solid
(iii) Milk – dispersion of liquid in liquid.
(iv) Paint – dispersion of solid in liquid.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Assertion and Reason Type

Question:73

In the following question, a statement of Assertion followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion An ordinary filter paper impregnated with collodion solution stops the flow of colloidal particles.
Reason (R): Pore size of the filter paper becomes more than size of colloidal particle.

(i) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (iii) Bigger size of colloidal particles do not allow the flow of colloidal particles through the smaller sized pores of filter paper.

Question:74

In the following question, a statement of Assertion followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion : Colloidal solutions show colligative properties.
Reason (R): Colloidal particles are large in size.

(i) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason arc correct but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (ii) The number of colloidal particles is lesser than the true solution since colloidal particles are large in size and leads to lower colligative properties.

Question:75

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): Colloidal solutions do not show Brownian motion.
Reason (R): Brownian motion is responsible for stability of sols.

(i) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason arc correct but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (v) Brownian movement is shown by colloidal particles and it is responsible for the stability of colloidal solution.

Question:76

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): Coagulation power of $\text {Al}^{3+}$ is more than $\text {Na}^{+}$
Reason (R): Greater the Valency of the flocculating ion added, greater is its power to cause precipitation (Hardy-Schulze rule).

(i) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason arc correct but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.

Answer:

The answer is the option (i) Hardy-Schulze law states that a higher valency of coagulating ion leads to higher power to coagulate the colloidal solution. So, $\text {Al}^{3+}$ has higher coagulation power than $\text {Na}^{+}$.

Question:77

In the following question, a statement of Assertion followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion :Detergents with low CMC are more economical to use.
Reason (R): Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC.

(i) Both Assertion and Reason are correct and the Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason arc correct but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (i) Clothes are cleaned by micellization, i.e. when concentration of detergents become equal to CMC. A lower CMC value means that the detergent is more economical.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 5: Long Answer Type

Question:78

What is the role of adsorption in heterogeneous catalysis?
Answer:

Heterogenous catalyst is used to increase the rate of reaction in which catalyst is not in phase with reactants and products. Gaseous molecules get adsorbed on the surface of the catalyst. These adsorbed molecules then dissociate, resulting in formation of active species which are more susceptible to reaction. Due to the high concentration of reactants, it the easily combines with other reactants or active species completing the chemical reaction. After the chemical reaction has taken place, products are desorped from the surface. New molecules can then occupy the freed up space.

Question:79

What are the applications of adsorption in chemical analysis?
Answer:

In chemical analysis, adsorption finds many applications. Some of these applications are:
(i) Qualitative analysis: Lake test which is used to confirm the presence of Aluminium ions is based on the principle of adsorption. Aluminum hydroxide can adsorb the blue colour of the litmus solution.
(ii) Adsorption indicators: For volumetric analysis, adsorption-based indicators are used. For instance, some dyes which adsorb on the precipitate can be used to figure out the end-point for the reaction.
(iii) Ion-exchange resins: Some organic polymers can selectively adsorb ions from water and are used to soften hard water.
(iv) Inert gas separation: Gases show a varying degree of adsorption by charcoal. Noble gases can be separated by adsorption by coconut charcoal at different temperatures.
(v) The froth floatation method used for concentration of sulphide ores is based on adsorption.
(vi) Chromatographic analysis: Adsorption chromatography is used to purify and separate pigments, hormones, etc.
(vii) In preparation of gas masks using activated charcoal to avoid poisonous gases.

Question:80

What is the role of adsorption in froth floatation process used especially for concentration of sulphide ores?

Answer:

The low grade sulphide ores are concentrated by froth floatation process. The ore is finely divided and is then put in pine oil and water mixture. Post this, compressed air is bubbled through the mixture. The foam will rise to the surface adsorbing ore particles and gangue particles will settle to the bottom.

Question:81

What do you understand by shape selective catalysis? Why are zeolites good shape-selective catalysts?
Answer:

Shape selective catalysis is a catalytic reaction, which distinguishes between the reactant, product or the transition state species in terms of relative sizes of molecules and pore space. Zeolites are excellent shape selective catalysis due to its honeycomb like structure. Zeolites, with a general formula $\text {M}_{\frac{\text {x}}{\text {n}}}[(\text {AlO}_{2})_{x}(\text {SiO}_{2})_{y}].\text {mH}_{2}\text {O},$ are microporous aluminosilicates. They can permit a molecule’s entry and exit into active regions and have a large inner surface area. In petrochemical industries, they find an application in isomerization and the cracking of hydrocarbons. The size of the cavities (caves) and pores (tunnels) determine the reaction’s feasibility. ZSM-5, a zeolite catalyst can convert alcohol to gasoline by dehydrating the alcohol in its cavities.


Class 12 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS) Questions

Question 1: 0.47 g of long chain fatty acid was dissolved in $100 \mathrm{~cm}^3$ of haxone, Hexane evaporates a monolayer is formed when 10 ml of this solution was added dropwise to the surface of water in round watch glass . The distance from edge to router of the watch glass is 10 cm . What is height of monolayer {Density of fatty acid $0.9 \mathrm{~g} \mathrm{~cm}^3 \pi=3$}

(1) $1.74 \times 10^{-4}$

(2) $1.86 \times 10^{-3}$

(3) $3.28 \times 10^{-2}$

(4) $2.48 \times 10^{-1}$

Solution: In 100 ml hexane $\longrightarrow 0.47 \mathrm{~g}$ fatty acid
10 ml of hexane $\longrightarrow 0.047 \mathrm{~g}$ fatty acid
Volum of fatty acid present in one round glass

$$
\begin{aligned}
& \frac{0.047}{0.9} \\
& \text { Area } \rightarrow \pi r^2 \\
& \text { Volume } \rightarrow \pi r^2 \times h \\
& \therefore \quad \pi r^2 * h=\frac{0.047}{0.9} \\
& 3 \times(10)^2 \times h=\frac{0.047}{0.9} \\
& \quad h=1.74 \times 10^{-4}
\end{aligned}
$$

Hence, the answer is the option (1).

Question 2: Which of the following is true?

(1) Animal hides are colloidal in nature.

(2) Addition of alum to water makes it unfit for drinking.

(3) Latex is not a colloidal solution.

(4) All of the above.

Solution: Applications of Colloids
1. Purification of Drinking water: Water obtained from natural sources contains suspended impurities. Alum is added to coagulate the suspended impurities and to make water fit for drinking.

2. Tanning: Animal hides are colloidal in nature. When a hide (positively charged) is soaked in tannin which contains negatively charged colloids, mutual coagulation occurs, and leather hardens.

3. Rubber industry: Latex is a colloidal solution of rubber particles which are negatively charged.

4. Formation of delta: These are prepared by coating an emulsion of the light sensitive silver bromide in gelatine over glass plates or celluloid films.

5. Industrial products: Paints, ink, synthetic plastics, rubber, graphite lubricants, cement etc.

Hence, the answer is the option (1).

Approaches to Solve Questions of Chapter 5 Surface Chemistry

To effectively solve questions from Chapter 5, Surface Chemistry, students can follow the approaches given below:

1) Understanding the concept of Adsorption

  • Finding the difference between Physisorption.
  • Adsorption isotherms
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2) Next step is Catalysis, where we must have the basic terms related to a catalyst, like:

  • Mechanism of catalysis
  • Type of catalysis
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3) Students must be aware of the Colloidal state, like:

  • Types of Colloids
  • Preparation of Colloids
  • Properties of Colloids
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4) The concept of Emulsion

  • The student must be aware of emulsion
  • Methods and preparation of Emulsion

5) Proper understanding of basic concepts and practice helps students clear their doubts and solve questions effectively.

Following are the major topics in Class 12 Chemistry NCERT Exemplar Chapter 5

  • Adsorption
  • Catalysis
  • Colloids
  • Classification of Colloids
  • Emulsions
  • Colloids Around Us

NCERT Exemplar Class 12 Chemistry Chapter 5 Surface Chemistry - Learning Outcome

This chapter has applications in many of the fields related to chemistry. The students will learn how to modify the chemical composition of a surface. And most importantly, the chapter has important usage in heterogeneous catalysis, electrochemistry, and geochemistry. So, the students will be learning many important aspects of chemistry through this chapter.

NCERT Exemplar Class 12 Chemistry Solutions Chapter-wise


Check NCERT Solutions for Class 12 Chemistry

Read more-

The NCERT subject-wise notes and solutions will help you in broadening your concept and will also help in revision. You can follow the links given in the table below. Also, you can find links to the syllabus and books recommended for the respective subjects.


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Frequently Asked Questions (FAQs)

1. What is Surface Chemistry?

Surface chemistry is the study of chemical reactions at surfaces and interfaces. It involves the interactions occurring between different phases, such as solid-gas, solid-liquid, and liquid-gas interfaces, and encompasses phenomena like adsorption, catalysis, and surface tension.


2. What is adsorption, and how is it different from absorption?

Adsorption is the process by which atoms, ions, or molecules from a gas, liquid, or dissolved solid adhere to a surface. In contrast, absorption involves the incorporation of a substance into the bulk of another material. Essentially, adsorption is a surface phenomenon, while absorption involves penetration of one substance into another.

3. What are the different types of adsorption?

There are two primary types of adsorption: physisorption and chemisorption. Physisorption is a physical process characterized by weak van der Waals forces acting between the adsorbate and the surface. At the same time, chemisorption involves the formation of chemical bonds and is generally stronger and more specific.

4. Can you explain the concept of adsorption isotherms?

Adsorption isotherms are curves that represent the relationship between the amount of adsorbate on the adsorbent and the concentration of the adsorbate in the surrounding phase at a constant temperature. Common models of adsorption isotherms include the Langmuir and Freundlich isotherms, which help predict how different conditions affect adsorption behavior.

5. How is surface tension defined in surface chemistry?

Surface tension is defined as the tendency of fluid surfaces to shrink into the minimum surface area possible. It is the result of cohesive forces between liquid molecules at the surface and is often measured in force per unit length (e.g., dynes/cm). It plays a critical role in various applications like emulsions and foams.


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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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