NCERT Solutions for Class 12 Science Chemistry Chapter 5

Question 1:

Which of the following process does not occur at the interface of phases?

(i) crystallisation
(ii) heterogenous catalysis
(iii) homogeneous catalysis
(iv) corrosion

Answer:

Homogenous catalysis does not occur at the interface of phases as all the reactants, products and catalysts are in same phase.
Hence, the correct answer is option (iii).

Question 2:

At the equilibrium position in the process of adsorption ___________.

(i) ∆H > 0
(ii) ∆H = T∆S
(iii) ∆H > T∆S
(iv) ∆H < T∆S

Answer:

We know that ∆G = ∆H − T∆S.
At equilibrium, ∆G = 0. So, ∆H = T∆S.
Hence, the correct answer is option (ii).

Question 3:

Which of the following interface cannot be obtained?

(i) liquid-liquid
(ii) solid-liquid
(iii) liquid-gas
(iv) gas-gas

Answer:

Air is a homogenous mixture of different gases. Therefore, gas-gas interface cannot be obtained.
Hence, the correct answer is option (iv).

Question 4:

The term ‘sorption’ stands for ____________.

(i) absorption
(ii) adsorption
(iii) both absorption and adsorption
(iv) desorption

Answer:

Sorption is the process when adsorption and absorption both occur simultaneously.
Hence, the correct answer is option (iii).

Question 5:

Extent of physisorption of a gas increases with ___________.

(i) increase in temperature.
(ii) decrease in temperature.
(iii) decrease in surface area of adsorbent.
(iv) decrease in strength of van der Waals forces.

Answer:

Physisorption of a gas is an exothermic process. So, according to Le-Chatelier's principle, it will occur at low temperature and its extent increases with decrease in temperature.
Hence, the correct answer is option (ii).

Question 6:

Extent of adsorption of adsorbate from solution phase increases with ________.

(i) increase in amount of adsorbate in solution.
(ii) decrease in surface area of adsorbent.
(iii) increase in temperature of solution.
(iv) decrease in amount of adsorbate in solution.

Answer:

The extent of adsorption of adsorbate from solution phase depends on concentration of adsorbate in solution and it increases with increase in amount of adsorbate in solution.
Hence, the correct answer is option (i).

Question 7:

Which one of the following is not applicable to the phenomenon of adsorption?

(i) ∆H > 0
(ii) ∆G < 0
(iii) ∆S < 0
(iv) ∆H < 0

Answer:

The phenomenon of adsorption is an exothermic process. Therefore, ∆H < 0.
Hence, the correct answer is option (i).

Question 8:

Which of the following is not a favourable condition for physical adsorption?

(i) high pressure
(ii) negative ∆H
(iii) higher critical temperature of adsorbate
(iv) high temperature

Answer:

Physical adsorption is an exothermic process and according to Le-Chatelier's principle, it is favoured at low temperature.
Hence, the correct answer is option (iv).

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 adsorbent
(iv) decrease in surface area of adsorbent

Answer:

Physical adsorption of gaseous species may change to chemisorption on increasing temperature as it increases the enthalpy of activation of adsorbate molecules. This leads to formation of stronger chemical bond.
Hence, the correct answer is option (ii).

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:

Physisorption has low specificity for any particular gas because it involves weak van der Waals' forces which are universal.
Hence, the correct answer is option (i).

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 metal surface

Answer:

Water is absorbed by calcium chloride as its concentration is uniform throughout the bulk of calcium chloride. All other options are examples of adsorption where concentration of adsorbate increases only on the surface of the adsorbent.
Hence, the correct answer is option (ii).

Question 12:

On the basis of data given below predict which of the following gases shows least adsorption on a definite amount of charcoal?
Gas CO₂ SO₂ CH₄ H₂
Critical temp./K 304 630 190 33

(i) CO₂
(ii) SO₂
(iii) CH₄
(iv) H₂

Answer:

Gases with higher critical temperatures are more easily adsorbed on the surface of charcoal as van der Waals' forces are stronger near the critical temperatures. H₂ has the least critical temperature of 33K among the given the options, so, it will show least adsorption on a definite amount of charcoal.
Hence, the correct answer is option (iv).

Question 13:

In which of the following reactions heterogenous catalysis is involved?

$(a) 2 {SO}_{2} (g) + O_{2} (g) \overset{NO (g)}{\to} {SO}_{3} (g) (b) 2 {SO}_{2} (g) \overset{Pt (s)}{\to} 2 {SO}_{3} (g) (c) N_{2} (g) + 3 H_{2} (g) \overset{Fe (s)}{\to} 2 {NH}_{3} (g) (d) {CH}_{3} {COOCH}_{3} (l) + H_{2} O (l) \overset{HCl (l)}{\to} {CH}_{3} COOH (aq) + {CH}_{3}^{} OH (aq)$

(i) (b), (c)
(ii) (b), (c), (d)
(iii) (a), (b), (c)
(iv) (d)

Answer:

Heterogenous catalysis is the catalysis in which reactants and catalysts are in different phases. In options (b) and (c), reactants are in gaseous phase while catalysts are in solid phase. So, they are examples of heterogenous catalysis.
Hence, the correct answer is option (i).

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:

At high concentration of soap in water, soap aggregate to form micelles which are also known as associated colloids.
Hence, the correct answer is option (ii).

Question 15:

Which of the following will show 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:

Tyndall effect is shown by colloidal particles and aqueous solution of soap above critical micelle concentration aggregates to form micelles which are also known as associated colloids.
Hence, the correct answer is option (ii).

Question 16:

Method by which lyophobic sol can be protected.

(i) By addition of oppositely charged sol.
(ii) By addition of an electrolyte.
(iii) By addition of lyophilic sol.
(iv) By boiling.

Answer:

Lyophobic sols can be protected by adding lyophilic sols. Lyophilic sols are more stable and form a layer around lyophobic sols to protect them.
Hence, the correct answer is option (iii).

Question 17:

Freshly prepared precipitate sometimes gets converted to colloidal solution by ___________.

(i) coagulation
(ii) electrolysis
(iii) diffusion
(iv) peptisation

Answer:

The process of converting freshly prepared precipitate to colloidal solution by shaking with dispersion medium in the presence of small amount of electrolyte is known as peptization.
Hence, the correct answer is option (iv).

Question 18:

Which of the following electrolytes will have maximum coagulating value for AgI/Ag⁺ sol?

(i) Na₂S
(ii) Na₃PO₄
(iii) Na₂SO₄
(iv) NaCl

Answer:

The coagulating power of anions follow the order: PO₄^3- > SO₄^2- > S^2- > Cl^-
The smaller the amount of oppositely charged ion needed, more is its coagulating power and value.
Hence, the correct answer is option (ii).

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:

A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is classified as sol. examples include paints and cell fluids.
Hence, the correct answer is option (iv).

Question 20:

The values of colligative properties of colloidal solution are of small order in comparison to those shown by true solutions of same concentration because of colloidal particles __________________.

(i) exhibit enormous surface area.
(ii) remain suspended in the dispersion medium.
(iii) form lyophilic colloids.
(iv) are comparatively less in number.

Answer:

The colloidal particles form aggregates and their number is comparatively less in solution. Therefore, the values of colligative properties of colloidal solution are of small order in comparison to those shown by true solutions of same concentration.
Hence, the correct answer is option (iv).

Question 21:

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

(i) a → b → c → d → e
(ii) a → c → b → d → e
(iii) a → c → b → e → d
(iv) a → b → c → e → d

Answer:

The mechanism of catalysis, in accordance with modern adsorption theory is as follows:

Diffusion of reactants to the surface of the catalyst.
Adsorption of reactant molecules on the surface of the catalyst.
Occurrence of chemical reaction on the catalyst’s surface through formation of an intermediate.
Desorption of reaction products from the catalyst surface, and thereby, making the surface available again for more reaction to occur.
Diffusion of reaction products away from the catalyst’s surface.

Hence, the correct answer is option (ii).

Question 22:

Which of the following process is responsible for the formation of delta at a place where rivers meet the sea?

(i) Emulsification
(ii) Colloid formation
(iii) Coagulation
(iv) Peptisation

Answer:

Coagulation is the process in which the electrolytes present in sea water coagulate the river water consisting of colloidal solution of clay to form a delta at a place where rivers meet the sea.
Hence, the correct answer is option (iii).

Question 23:

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

Answer:

Freundlich adsorption isotherm can be expressed by the following equation:
$\frac{x}{m} = k . p^{\frac{1}{n}} (n > 1)$
On taking logarithm,
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
This equation can be compared to straight line equation y = mx + c
$y = \log \frac{x}{m} S l o p e = m = \frac{1}{n} x = \log p I n t e r c e p t = c = \log k$
Hence, the correct answer is option (iii).

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:

Absorption of ionic species from solution is not responsible for the presence of electric charge on the sol particles.
Hence, the correct answer is option (iv).

Question 25:

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

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

Answer:

The given figure illustrates the process of adsorption of colouring substances by animal charcoal due to which solution of raw sugar becomes colourless.
Hence, the correct answer is option (ii).

Question 26:

In the given question two or more options may be correct.

Which of the following options 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 as a normal strong electrolyte at all concentrations.

Answer:

Soap solution behave as strong electrolyte(ion) at low concentration and aggregate to form micelles above a particular temperature called Kraft temperature(T_k) and above a certain concentration known as critical micelle concentration(CMC). On dilution, micelles revert to individual ions as their concentration decreases.
Hence, the correct answers are options (ii) and (iii).

Question 27:

In the given question two or more options may be correct.

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 ∆H of reaction.
(iii) Catalyst is required in large quantities to catalyse reactions.
(iv) Catalytic activity of a solid catalyst does not depend upon the strength of chemisorption.

Answer:

Solid catalyst is highly selective and yields a particular product for a particular reaction. Different catalysts give different products under same reaction conditions. It does not change ∆H of reaction but only lowers the activation energy without being consumed in the reaction. It is needed in small amounts and its activity depends on the strength of chemisorption.
Hence, the correct answers are options (i) and (ii).

Question 28:

In the given question two or more options may be correct.

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

Answer:

When $\frac{1}{n} = 0$ , the expression for Freundlich adsorption isotherm becomes
$\frac{x}{m} = k p^{0} = k . 1 = k$
Thus, the adsorption is independent of pressure.

When n = 0, the expression for Freundlich adsorption isotherm becomes
$\frac{x}{m} = k p^{\frac{1}{0}} = k p^{\infty}$
Thus, $\frac{x}{m}$ vs p graph will be a line parallel to x-axis.

Hence, the correct answers are options (i) and (iii).

Question 29:

In the given question two or more options may be correct.

H₂ 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:

H₂ gas has a very low critical temperature and very weak van der Waals' forces. Therefore, it is adsorbed on activated charcoal to a very little extent in comparison to easily liquefiable gases.
Hence, the correct answers are options (ii) and (iii).

Question 30:

In the given question two or more options may be correct.

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 mixing of two oppositely charged sols neutralizes their charges and cause coagulation of sols. The presence of equal and similar charges prevent colloidal particles to aggregate due to repulsive forces and stabilize the sols. The addition of dispersed liquid to an emulsion creates a separate layer. Brownian movement stabilizes sols by a stirring effect which does not allow colloidal particles to settle down.
Hence, the correct answers are options (ii) and (iv).

Question 31:

In the given question two or more options may be correct.

An emulsion cannot be broken by __________ and ___________.

(i) heating
(ii) adding more amount of dispersion medium
(iii) freezing
(iv) adding emulsifying agent

Answer:

An emulsion can be broken into its liquid components by heating, freezing, centrifuging, etc. But addition of any amount of dispersion medium and emulsifying agent does not break an emulsion rather stabilizes it.
Hence, the correct answers are options (ii) and (iv).

Question 32:

In the given question two or more options may be correct.

Which of the following substances will precipitate the negatively charged emulsions?

(i) KCl
(ii) glucose
(iii) urea
(iv) NaCl

Answer:

The negatively charged emulsions are precipitated by addition of electrolytes like KCl and NaCl.
Hence, the correct answers are options (i) and (iv).

Question 33:

In the given question two or more options may be correct.

Which of the following colloids cannot be coagulated easily?

(i) Lyophobic colloids.
(ii) Irreversible colloids.
(iii) Reversible colloids.
(iv) Lyophilic colloids.

Answer:

Lyophilic colloids are quite stable due to the charge present on them and solvation property of colloidal particles. Moreover, when dispersion medium is separated from dispersed phase, they can revert back to their colloidal form on remixing of dispersion medium. Therefore, they are also called reversible colloids.
Hence, the correct answers are options (iii) and (iv).

Question 34:

In the given question two or more options may be correct.

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:

Lyophobic sols can be protected by adding lyophilic sols. Lyophilic sols are more stable and form a layer around lyophobic sols to protect them.
Hence, the correct answers are options (i) and (iii).

Question 35:

In the given question two or more options may be correct.

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:

Dispersion medium begins to move and electroosmosis takes place when an electric field is applied to a colloidal solution and electrophoresis is prevented.
Hence, the correct answers are options (ii) and (iii).

Question 36:

In the given question two or more options may be correct.

In a reaction, catalyst changes ____________.

(i) physically
(ii) qualitatively
(iii) chemically
(iv) quantitatively

Answer:

A catalyst accelerates the rate of a reaction without being consumed in the reaction chemically and quantitatively. Thus, it changes physically and qualitatively only.
Hence, the correct answers are options (i) and (ii).

Question 37:

In the given question two or more options may be correct.

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) absoprtion of solvent

Answer:

When a chalk stick is dipped in ink, the coloured substance is adsorbed on the surface while solvent is absorbed deeper into the stick.
Hence, the correct answers are options (i) and (iv).

Question 38:

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

Answer:

The objective of surface studies is to facilitate adsorption of desired substances on surface of adsorbents. Therefore, it is essential to have clean surfaces so that maximum surface area is available for adsorption to take place.

Question 39:

Why is chemisorption referred to as activated adsorption?

Answer:

Chemisorption requires high activation energy for formation of strong chemical bonds. Hence, it is also known as activated adsorption.

Question 40:

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

Answer:

Soap solution behave as strong electrolyte(ion) at low concentration in water and aggregate to form micelles above a certain concentration known as critical micelle concentration(CMC).

Question 41:

What happens when gelatin is mixed with gold sol?

Answer:

Gold sol is lyophobic and it gets stabilized when gelatin, a lyophilic colloid, is mixed with it. Gelatin forms a protective sheath around gold sol and protects it from electrolytes. This is an example of protective action of lyophilic colloids.

Question 42:

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

Answer:

Clouds are charged aerosols - colloidal solutions of rain droplets suspended in air. When silver iodide is sprayed on them, they get coagulated due to neutralization of charge and result in rain which is known as artificial 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 when added in ice-creams. Ice cream is an oil in water type emulsion and gelatin is added to stabilize this emulsion.

Question 44:

What is collodion?

Answer:

Colloidon is a 4% solution of nitrocellulose in a mixture of alcohol and ether. Normal filter paper is soaked in collodion solution to reduce the size of pores of filter paper and convert it into ultra filter paper. This ultra filter paper can then be used to filter colloidal particles.

Question 45:

Why do we add alum to purify water?

Answer:

Alum (phitkari) is added to impure water for coagulation of suspended impurities present in water. After coagulation, impurities settle down and pure water is obtained.

Question 46:

What happens when electric field is applied to colloidal solution?

Answer:

Colloidal particles contain charge and when electric field is applied to solution containing colloidal particles, these particles move towards oppositely charged electrodes. Negatively charged particles move towards the anode while positively charged particles migrate towards the cathode.

Question 47:

What causes brownian motion in colloidal dispersion?

Answer:

The unbalanced bombardment of dispersed phase particles by dispersion medium molecules is the main cause of Brownian motion in colloidal dispersion. It has a stirring effect which does not allow colloidal particles to settle down and leads to stabilization of colloidal dispersion.

Question 48:

A colloid is formed by adding FeCl₃in excess of hot water. What will happen if excess sodium chloride is added to this colloid?

Answer:

On addition of FeCl₃in excess of hot water, a positively charged colloid of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.
Fe₂O₃.xH₂O/Fe³⁺
When an electrolyte like sodium chloride is added to this positively charged colloid, negatively charged Cl^- ions neutralize the positive charge and cause coagulation of the colloid.

Question 49:

How do emulsifying agents stabilise the emulsion?

Answer:

Sometimes, emulsions separate into two layers when left undisturbed. They are stabilized by adding an emulsifying agent which forms an interfacial film between suspended particles and the medium.

Question 50:

Why are some medicines more effective in the colloidal form?

Answer:

The effectiveness of medicines depends on how easily they are assimilated in the body and medicines in colloidal form are assimilated more easily owing to their large surface area. Therefore, some medicines are more effective in the colloidal form.

Question 51:

Why does leather get hardened after tanning?

Answer:

Tannin contains negatively charged colloidal particles. When animal hides containing positively charged particles are soaked in it, mutual coagulation takes place between the two colloidal sols. This process is called tanning and leads to hardening of leather.

Question 52:

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

Answer:

Cottrell precipitator is a chamber through which colloidal smoke particles are passed before they come out of the chimney. The chamber has plates containing charges opposite to that of smoke particles. When these particles come in contact with the plates, they lose their charge and get precipitated.

Question 53:

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

Answer:

An emulsion can be diluted to any extent without compromising with its stability when dispersion medium is added to it. On the contrary, a separate layer is formed after the mixing of dispersed phase. In this way, we can distinguish between dispersed phase and dispersion medium in an emulsion.

Question 54:

On the basis of Hardy-Schulze rule explain why the coagulating power of phosphate is higher than chloride.

Answer:

According to Hardy-Schulze rule, greater the charge on flocculating ion, lesser will be its amount needed to cause precipitation and more will be its coagulation power. Therefore, in case of coagulation of a positive sol, phosphate (PO₄^3-) with −3 charge will have more coagulating power than chloride (Cl^-) with −1 charge.

Question 55:

Why does bleeding stop by rubbing moist alum?

Answer:

Blood is a colloid containing albumin proteins. When moist alum (phitkari) is rubbed, bleeding stops as charged proteins of blood are coagulated by oppositely charged electrolytic ions present in alum.

Question 56:

Why is Fe(OH)₃colloid positively charged, when prepared by adding FeCl₃to hot water?

Answer:

On addition of FeCl₃in excess of hot water, a positively charged colloid of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.
Fe₂O₃.xH₂O/Fe³⁺

Question 57:

Why do physisorption and chemisorption behave differently with rise in temperature?

Answer:

Physisorption involves weak van der Waals' forces between adsorbate and adsorbent which decrease with increase in temperature. On the other hand, chemisorption involves strong chemical bonds which require activation energy and are favoured with rise in temperature. Therefore, physisorption and chemisorption behave differently with rise in temperature.

Question 58:

What happens when dialysis is prolonged?

Answer:

Dialysis involves removal of dissolved substances from a colloidal solution via diffusion through suitable membrane. But prolonged dialysis i.e. dialysis for a long time results in removal of essential electrolytes that stabilize colloidal solution and thus, it becomes unstable.

Question 59:

Why does the white precipitate of silver halide become coloured in the presence of dye eosin.

Answer:

Eosin dye is adsorbed on the surface of white precipitate of silver halide and therefore, the precipitate becomes coloured.

Question 60:

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

Answer:

Activated charcoal performs the role of adsorbent for adsorbing poisonous gases in coal mines so that the workers do not inhale these poisonous gases.

Question 61:

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

Answer:

When river water, which is a colloidal solution of clay meets the sea water, the ions present in sea water coagulate the colloidal clay solution. This results in its deposition and delta is formed.

Question 62:

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

Answer:

At low temperature, dihydrogen is adsorbed on nickel surface by van der Waals’ forces. But as the temperature is increased, molecules of hydrogen dissociate to form hydrogen atoms which are held on the surface by strong chemical bonds. Thus, a physical adsorption at low temperature may change into chemisorption when the temperature is increased.

Question 63:

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

Answer:

Desorption is important as the surface of the catalyst is again free for reaction with new substrate molecules. Thus, it helps in increasing the efficiency of a catalyst.

Question 64:

What is the role of diffusion in heterogenous catalysis?

Answer:

Diffusion is important for bringing in reactant molecules to the catalyst surface and removing product molecules from the catalyst surface. Thus, it helps in freeing up the vacant sites of adsorption of new reactant molecules.

Question 65:

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

Answer:

A solid catalyst enhances the rate of combination of gaseous molecules by bringing them in contact with the surface of a solid catalyst. As a result, a weak chemical combination takes place between the gaseous molecules and the surface of the catalyst which increases the concentration of reactants on the surface. When they are adsorbed side by side, they have better chance to react and form new molecules.

Question 66:

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

Answer:

Yes, the vital functions of the body such as digestion get affected during fever.
This is because during fever, the temperature of the body is increased above the normal temperature. As a result, the activity of the enzymes get affected because they work in an optimal range of 298 K - 310 K.

Question 67:

Match the items of Column I and Column II in the given question.

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 vapours 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:

Column I	Column II
(i) Sulphur vapours passed through cold water	(b) Molecular colloids
(ii) Soap mixed with water above critical micelle concentration	(c) Associated colloid
(iii) White of egg whipped with water	(d) Macro molecular colloids
(iv) Soap mixed with water below critical micelle concentration	(a) Normal electrolyte solution

Question 68:

Match the items of Column I and Column II in the given question.

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 semi permeable 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 semi permeable membranes towards solution side	(d) Reverse osmosis

Answer:

Column I	Column II
(i) Dispersion medium moves in an electric field	(c) Electroosmosis
(ii) Solvent molecules pass through semi permeable membrane towards solvent side	(d) Reverse osmosis
(iii) Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes	(b) Electrophoresis
(iv) Solvent molecules pass through semi permeable membranes towards solution side	(a) Osmosis

Question 69:

Match the items of Column I and Column II in the given question.

Match the items given in Column I and Column II.

Column I	Column II
(i) Protective colloid	(a) FeCl₃+ NaOH
(ii) Liquid - liquid colloid	(b) Lyophilic colloids
(iii) Positively charged colloid	(c) Emulsion
(iv) Negatively charged colloid	(d) FeCl₃+ hot water

Answer:

Column I	Column II
(i) Protective colloid	(b) Lyophilic colloids
(ii) Liquid - liquid colloid	(c) Emulsion
(iii) Positively charged colloid	(d) FeCl₃+ hot water
(iv) Negatively charged colloid	(a) FeCl₃+ NaOH

Question 70:

Match the items of Column I and Column II in the given question.

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 in liquid	(d) Emulsion

Answer:

Column I	Column II
(i) Solid in liquid	(b) Sol
(ii) Liquid in solid	(c) Gel
(iii) Liquid in liquid	(d) Emulsion
(iv) Gas in liquid	(a) Foam

Question 71:

Match the items of Column I and Column II in the given question.

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:

Column I	Column II
(i) Dialysis	(d) Purification
(ii) Peptisation	(c) Colloidal sol formation
(iii) Emulsification	(a) Cleansing action of soap
(iv) Electrophoresis	(b) Coagulation

Question 72:

Match the items of Column I and Column II in the given question.

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:

Column I	Column II
(i) Butter	(d) dispersion of liquid in solid
(ii) Pumice stone	(c) dispersion of gas in solid
(iii) Milk	(a) dispersion of liquid in liquid
(iv) Paints	(b) dispersion of solid in liquid

Question 73:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : An ordinary filter paper impregnated with collodion solution stops the flow of colloidal particles.
Reason : Pore size of the filter paper becomes more than the size of colloidal particle.

(i) Assertion and reason both are correct and the reason is correct explanation of assertion.
(ii) Assertion and reason both are correct but reason does not explain 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:

Colloidal particles can easily pass through an ordinary filter paper because the pores of the filter paper are larger in size than the size of colloidal particles but when it is impregnated with collodion solution, the pore size decreases and hence, the colloidal particles do not pass through it and get separated from mixture.
Hence, the correct answer is option (iii).

Question 74:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Colloidal solutions show colligative properties.
Reason : Colloidal particles are large in size.

(i) Assertion and reason both are correct and the reason is correct explanation of assertion.
(ii) Assertion and reason both are correct but reason does not explain 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:

Colloidal particles are large in size and therefore, the number of particles in colloidal solutions are less than that in true solutions. So, they show colligative properties of small order.
Hence, the correct answer is option (ii).

Question 75:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Colloidal solutions do not show brownian motion.
Reason : Brownian motion is responsible for stability of sols.

(i) Assertion and reason both are correct and the reason is correct explanation of assertion.
(ii) Assertion and reason both are correct but reason does not explain 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:

Colloidal particles undergo a continuous random zig-zag motion when viewed under a powerful ultramicroscope. This motion is known as Brownian motion. It has a stirring effect which does not allow the particles to settle and thus, is responsible for the stability of sols.
Hence, the correct answer is option (v).

Question 76:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Coagulation power of Al³⁺ is more than Na⁺.
Reason : Greater the valency of the flocculating ion added, greater is its power to cause precipitation (Hardy Schulze rule).

(i) Assertion and reason both are correct and the reason is correct explanation of assertion.
(ii) Assertion and reason both are correct but reason does not explain 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 coagulation power of Al³⁺ is more than Na⁺because it has more positive charge than Na⁺. According to Hardy Schulze rule, greater the valency of the flocculating ion added, greater is its power to cause precipitation.
Hence, the correct answer is option (i).

Question 77:

In the given question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion : Detergents with low CMC are more economical to use.
Reason : Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC.

(i) Assertion and reason both are correct and the reason is correct explanation of assertion.
(ii) Assertion and reason both are correct but reason does not explain 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 cleansing action of detergents involves the formation of micelles which are formed when the concentration of detergents becomes equal to CMC. Therefore, detergents with low Critical Micelle Concentration (CMC) are more economical to use.
Hence, the correct answer is option (i).

Question 78:

What is the role of adsorption in heterogenous catalysis?

Answer:

Adsorption of reactant molecules on the solid surface of catalysts increases the rate of reaction in heterogenous catalysis. The concentration of reactant molecules increases on the surface of the catalyst after being adsorbed. As a result, chemical combination between them increases because they are present side by side to each other.

Question 79:

What are the applications of adsorption in chemical analysis?

Answer:

The applications of adsorption in chemical analysis are as follows:

Chromatographic analysis: Chromatographic techniques like Thin Layer Chromatography and Paper Chromatography are based on the principle of adsorption and find a number of applications in industrial and analytical fields.
Adsorption indicators: Some dyes like eosin, fluorescein, etc. are adsorbed on the surfaces of certain precipitates such as silver halides and a characteristic colour of precipitate is obtained at the end point.
Qualitative analysis: Confirmation tests of some basic radicals such as lake test for Al⁺³ require the use of adsorption. The adsorption of colour from blue litmus by the precipitate of Al(OH)₃ confirms the presence of Al⁺³ ion.

Question 80:

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

Answer:

Sulphide ore is concentrated by separating it from silica and other impurities by froth floatation process using pine oil and frothing agent. A suspension of powdered ore is formed with water and pine oil is added to it. The mineral particles are wetted by pine oil while the impurities by water. When the mixture is agitated, the mineral particles come up with froth and are skimmed off. Then, the sulphide ores are recovered from froth with the help of drying.

Question 81:

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

Answer:

Shape-selective catalysis is a type of catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactants and products.
Zeolites are are microporous three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving Al–O–Si framework. They are good shape-selective catalysts because they have honeycomb-like structures and the reactions taking place in them depend upon the size and shape of reactants and products as well as upon the pores and cavities of the zeolites.

NCERT Solutions for Class 12 Science Chemistry Chapter 5 - Surface Chemistry

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