NCERT Solutions for Class 12 Chemistry Chapter 5 - Surface Chemistry

Question 1:

Write any two characteristics of chemisorption.

Answer:
  1. The attractive forces between adsorbent and adsorbate molecules are strong chemical bonds and therefore, Molar enthalpy of adsorption is high (of the order of 80 – 240 kJ mol–1).
  2. It is irreversible in nature.
Question 2:

Why does physisorption decrease with the increase of temperature ?

Answer:

In physisorption, the attractive forces between adsorbent and adsorbate molecules are weak van der Waal’s forces. When temperature is increased, the kinetic energy of the molecules of the gas increases and they can easily leave the surface of adsorbent because of weak forces of attraction.

Question 3:

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

Answer:

This is because powdered substances have more surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.

Question 4:

n Haber’s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming.

Answer:

Carbon monoxide acts as a poison for the catalyst in Haber process and therefore, it will lower the activity of the catalyst. Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 5:

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

Answer:

This is because of the process of autocatalysis. Ester on hydrolysis gives an acid which starts acting as a catalyst after sometime and therefore, the reaction becomes fast.

Question 6:

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

Answer:

In the process of catalysis, when desorption occurs from the surface of the catalyst, the surface becomes ready to adsorb fresh reactants and act as a catalyst.

Question 7:

What modification can you suggest in Hardy Schulze law ?

Answer:

The Hardy Schulze law considers the coagulation of sols., because of neutralization of their charges. Since coagulation can also occur by mixing two oppositely charged sols., it should also include.

“When oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, the coagulation of both the sols occurs”.

Question 8:

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

Answer:

The precipitates may be contaminated with the adsorbed particles of some other substances (impurities) which might have been used or formed during precipitation. Therefore, it is essential to wash the precipitate to remove these adsorbed particles.

Question 9:

Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.

Answer:

Adsorption is a phenomenon in which there is higher concentration of another substance on the surface than in the bulk. Absorption is a phenomenon in which the molecules of a substance are uniformly distributed throughout the body of the other substance.

For example, silica gel adsorbs water vapour while anhydrous calcium chloride absorbs water.

Question 10:

What is the difference between physisorption and chemisorption ?

Answer:

(i) Physisorption. When the particles of the adsorbate are held to the surface of the adsorbent by the physical forces such as van der Waal’s forces, the adsorption is called physical adsorption or physisorption. The attractive forces are weak and, therefore, these can be easily overcomed either by increasing the pressure or by decreasing the pressure. In other words, physical adsorption can be easily reversed or decreased.

(ii) Chemisorption. When the molecules of the adsorbate are held to the surface of the adsorbent by the chemical forces, the adsorption is called chemical adsorption or chemisorption. In this case, a chemical reaction occurs between the adsorbed molecules and the molecules or atoms of adsorbent.

Question 11:

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

Answer:

Finely divided substance has larger surface area and hence greater adsorption.

Question 12:

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

Answer:

The adsorption of a gas on a solid depends upon the following factors :

(i) Nature of the gas. The adsorption depends upon the nature of the gas adsorbed. The easily liquefiable gases such as HCl, NH3, Cl2,etc. are adsorbed more than the permanent gases such as H2, N2 and O2. However, in the case of chemical adsorption a gas gets adsorbed on the solid only if it forms chemical bonds.

(ii) Nature of adsorbate. Activated charcoal can adsorb gases which are easily liquefied. Gases such as H2, N2 and O2 are generally adsorbed on finely divided transition metals e.g., Ni and Co. The extent of adsorption depends on the available surface.

(iii) Effect of pressure. The extent of adsorption of a gas per unit mass of adsorbent depends upon the pressure of the gas. The variation of extent of adsorption (expressed as x/m where x is the mass of adsorbate and m is the mass of the adsorbent and the pressure is given in Fig. A graph between the amount of adsorption and gas pressure keeping the temperature constant is called an adsorption isotherm.

It is clear from the figure that extent of adsorption (x/m) increases with pressure and becomes maximum corresponding to pressure ps called equilibrium pressure. At this pressure the extent of adsorption becomes constant even though the pressure is increased. This state is also called saturation state and ps is called saturation pressure. The variation of extent of adsorption (x/m) with pressure (p) was given mathematically by Freundlich as :

where n is a whole number. This relationship is known as Freundlich adsorption isotherm.

(iv) Effect of temperature. Since the process of adsorption is exothermic, the rate of adsorption will not be favoured by the increase in temperature. With the increase in temperature at constant pressure, the extent of adsorption (x/m) will decrease as shown in Fig. (a). The graph between extent of adsorption and temperature at constant pressure is called adsorption isobar.

The adsortion isobars for physical and chemical adsorption are different. In physical adsorption, there is a regular decrease as temperature increases. However, in chemisorption, there is initial increase and then it decreases. This is shown in Fig. (b).

(v) Activation of adsorbent. Activation of adsorbent means increasing of the adsorption power of the adsorbent. It is very necessary to increase the rate of adsorption. The can be done by the following methods :

  1. Metallic adsorbents are activated by mechanical rubbing or by subjecting it to some chemical reactions.
  2. To increase the adsorbing power of adsorbents, they are sub-divided into smaller pieces. As a result, the surface area increases and therefore, the adsorbing power increases.
  3. Some adsorbents are activated by strong heating in contact with superheated steam. For example, charcoal is activated by subjecting it to the action of superheated steam.
Question 13:

What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Answer:

Adsorption isotherm. A graph between amount of adsorption and gas pressure keeping temperature constant is called an adsorption isotherm.
Freundlich adsorption isotherm. The behaviour of adsorption with pressure can be expressed by adsorption isotherm as shown in figure. The inspection of the graph reveals the following facts :

(i) At low pressure, the graph is almost straight line which indicates that x/m is directly proportional to the pressure.
This may be expressed as :

where k is constant.

(ii) At high pressure. The graph becomes almost constant which means that x/m becomes independent of pressure. This may be expressed as :

(iii) Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between 0 to 1 i.e., fractional power of pressure. This may be expressed as :

where n can take any value between 1 and a larger integer depending upon the pressure. The above relationship is also called Freundlich’s adsorption isotherm.

Question 14:

What role does adsorption play in heterogeneous catalysis ?

Answer:

In heterogeneous catalysis, the reactants are generally gases while catalysts are solids. The reactant molecules are adsorbed on the surface of the solid catalyst by physical adsorption or chemical adsorption. As a result, the concentration of the reactant molecules on the surface of the catalyst increases and hence the rate of reaction also increases.

Alternatively, one of the reactant molecules undergo fragmentation on the surface of the catalyst producing active sites which make the reaction fast. The product molecules, on the other hand, have no affinity for the solid catalyst and therefore, undergo desorption leaving the surface free for further fresh adsorption. This theory is called adsorption theory.

Question 15:

Why is adsorption always exothermic ?

Answer:

Adsorption occurs because of attraction between adsorbate and adsorbent molecules and therefore, energy is always released during adsorption. Hence adsorption is an exothermic process.

Question 16:

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

Answer:

Colloids consists of two phases ; a dispersion medium and a dispersed phase. The component present in excess (just like a solvent in a solution) is termed as a dispersion medium while the component present in small proportion (just like a solute in a solution) is called the dispersed phase. Out of the three phases of matter solid, liquid or gas each one can act as a dispersed phase or a dispersed medium giving eight colloidal systems. These are listed below:

Based on the nature of dispersion media, colloids in water are sometimes called hydrosols, those in alcohols are alcosols and those in benzene are benzosols.

Question 17:

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

Answer:

The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, solutions of metals like Ag, Au, Al(OH)3, Fe(OH)3, etc.

The lyophobic sols are less stable because their stability is due to charge only. If the charge is removed, the particles will come nearer to each other to form aggregates i.e., they will coagulate and settle down. On the other hand, the stability of lyophilic sols is due to charge as well as solvation of colloidal particles. Therefore, these are not easily coagulated.

Question 18:

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

Answer:

The important differences between multimolecular and macromolecular colloids are:

The common example of multimolecular colloid is sulphur sol, which consists of particles of S8 molecules. The common example of macromolecules is starch.

The associated colloids are substances which behave as normal electrolytes at low concentrations but behave as colloidal particles at higher concentration. For example, soap and detergents.

They differ from multimolecular and macromolecular colloids in the sense that they behave as normal electrolytes at low concentrations but exhibit colloidal state properties at higher concentrations due to the formation of aggregated particles called micelles.

Question 19:

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

Answer:

Enzymes are complex nitrogeneous organic compounds produced by living cells which catalyze the biochemical reactions occurring in living organisms. These are also called biological catalysts or bio-chemical catalysts. Without enzymes, the living process would be very slow to sustain life. All enzymes are proteins. About 3000 enzymes have been identified.

The enzymes increase the speed of reactions upto 10 million times as compared to the uncatalysed reactions. These are highly specific in nature. Almost every biochemical reaction is controlled by its own specific enzymes.

Mechanism of enzyme catalysts

The various steps involved in the enzyme catalysed reaction is shown in Fig.1.

  1. Binding of the enzyme (E) to substrate (S) to form a complex.
    E + S ES
    ES is called the enzyme-substrate complex.
  2. Product formation in the complex.
    ES EP
    where EP is a complex of enzyme and product.
  3. Release of product from the enzyme-product complex.
    EP E + P
    The catalytic property of enzymes is present at certain specific regions on their surfaces. These are called active sites or catalytic sites. The active sites have characteristic shape and fit suitably shaped specific areas on the surface of the substrate molecules. Specific binding accounts for the high specificity of these enzymatic reactions. The enzyme reaction may be explained in terms of lock and key model. According to this model, the substrate, the molecule on which the enzyme acts, fits into the slot as key fits into a lock. The shape of the active site of any given enzyme is such that only a specific substrate can fit into it, in the same way as one key can open a particular lock as shown below :
Question 20:

How are colloids classified on the basis of (a) physical states of components, (b) nature of dispersed phase, and (c) interaction between dispersed phase and dispersion medium ?

Answer:
  1. Classification of colloids on the basis of physical states of components. The colloids can be classified on the basis of physical state of dispersion medium and dispersed phase. For detail, Refer Textbook Exercises; Q.9.
  2. Classification on the basis of nature of dispersed phase. Refer Textbook Exercises; Q.9.
  3. Classification on the basis of nature of interactions between dispersed phase and dispersion medium. These may be classified as
    1. lyophilic sols and
    2. lyophobic sols.

Refer Textbook Exercises; Q.11.

Question 21:

Explain what is observed
(i) when a beam of light is passed through a colloidal sol., (ii) an electrolyte, NaCl is added to hydrated ferric oxide sol. (iii) electric current is passed through a colloidal sol. ?

Answer:
  1. Scattering of light by colloidal particles takes place and path of light becomes visible (Tyndall effect).
  2. The positively charged colloidal particles of Fe(OH)3 get coagulated by the oppositely charged Cl ions provided by NaCl.
  3. On passing electric current, the colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated. This is electrophoresis process.
Question 22:

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

Answer:

Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium.

Types of emulsion. These are of two types :

  1. Oil-in water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess) e.g., milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.
  2. Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium e.g., butter, cod liver oil, cold cream etc. Such types of emulsions are called oily emulsions.
Question 23:

How do emulsifiers stabilise emulsion? Name two demulsifiers.

Answer:

To stabilise emulsion, small quantities of certain substances are added, called emulsifiers. These reduce the interfacial tension between the two liquids forming the emulsion. The common emulsifiers are gum, gelatin, albumin, etc.

Question 24:

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

Answer:

Soap stabilises the emulsion between grease and water by forming micelles around the grease (and dirt). This is the cleansing action of soap.
Soaps or detergents have been used as cleaning agents. The soaps are sodium salts of long chain fatty acids. They have a polar end COO– Na+ and non-polar end made of long chain hydrocarbons which may be —C15H31, —C17H33, —C17H35 etc. For example, sodium stearate is composed of long chain of alkyl group called tail and a polar part of carboxylate ion and sodium ion called head.

This may be represented as the polar end which is water soluble and the non-polar end

which is soluble in organic solvents and oil. Ordinarily, the dirt in the cloth is due to the presence of dust particles in fat or grease which stick to the cloth. When the cloth is dipped in aqueous soap solution, the soap and the dirt come in contact with

each other. The non-polar end (tail) is directed towards the oil or grease present on the cloth and the polar end (head) is directed towards water. In this manner, each oil droplet is surrounded by a number of negatively charged carboxylate ions. Since similar charges repel each other, the oil droplets are prevented from coming in contact with each other. As a result, oil droplets break up and more small droplets are formed. The hand rubbing or the agitation due to the washing machine causes dispersion of the oil or grease throughout the soapy water. In this way grease or dirt are removed from the surface of the cloth.

Question 25:

Give four examples of heterogeneous catalysis.

Answer:

Examples of heterogeneous catalysis :

  1. Manufacture of ammonia by Haber’s process using iron catalyst.

  2. Manufacture of SO3 from SO2 in the Contact process using V2O5 as catalyst.

  3. Dehydrogenation of ethanol by Ni catalyst

  4. Reduction of nitrobenzene in the presence of Ni as catalyst.

Question 26:

What do you mean by activity and selectivity of catalysts ?

Answer:

Activity means the ability of catalysts to increase the chemical reaction. A common example of activity is the reaction H2 and O2 to form water in the presence of platinum. Without the presence of the catalyst, the mixture of H2 and O2 can be safely stored. But in the presence of catalyst, the reaction occurs with explosive voilence. In some cases the catalyst can accelerate the reaction to as high as 108 times.
The selectivity means the ability of the catalyst to direct reaction to give particular products. For example, n-heptane in the presence of platinum catalyst gives toluene.

Similarly, propylene (CH3CH==CH2) reacts with O2 in the presence of bismuth molybdate catalyst to selectively give acrolein (CH2==CHCHO)

Question 27:

Describe some features of catalysis by zeolites.

Answer:

Zeolites are microporous aluminosilicates of the general formula Mx/n[(AlO2)x (SiO2)y]mH2O. These are most important oxide catalysts. These are used in petrochemical industries for cracking of hydrocarbons and isomerization. The reactions in zeolites depend upon the size of the cavities (cages) or pores (apertures) present in them. The most remarkable feature of zeolite catalysis is the shape selectivity. Therefore, the selectivity of catalyst depends on the pores structure. It has been observed that the pores size in zeolites generally varies between 260 pm to 740 pm. Depending upon the size of the reactants and products compared to the size of the cages or pores of zeolite, reactions proceed in specific manner.

A zeolite catalyst called ZSM-5 converts alcohols to gasoline, by first dehydrating the alcohol by loss of water.

Question 28:

Explain the following terms :
(i) Electrophoresis (ii) Coagulation (iii) Dialysis (iv) Tyndall effect.

Answer:

(i) Electrophoresis. The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. This helps to know the presence of charge on the sol particles and its nature whether positive or negative. In this experiment, the colloidal particles move towards positive or negative electrodes depending upon their charge under the influence of electrical field. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative. The apparatus consists of U-tube with two platinum electrodes in each limb as shown in Fig. Take a sol of As2S3 in the U-tube. The intensity of the colour of the sol in both the arms is same. Now, pass the current through the sol. After sometime, it is observed that the colour of the sol near the positive electrode became intense than the initial colour. This indicates that the As2S3 particles have accumulated near the positive electrode. Therefore, the particles of As2S3 are negatively charged and they move towards oppositely charged (positive) electrode and get accumulated there.

Similarly, when an electric current is passed through positively charged Fe(OH)3 sol, it is observed that they move towards negatively charged electrode and get accumulated there. Thus, by observing the direction of movement of the colloidal particles, the sign of the charge carried by the particles can be determined.

(ii) Coagulation. The phenomenon of precipitation of a colloidal solution by the addition of excess of an electrolyte is called coagulation or flocculation. The coagulation capacity of different electrolytes depends upon the valency of the active ion or called flocculating ion. It is the ion carrying charge opposite to the charge on the colloidal particles. According to Hardy Schulz law, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. Thus, to coagulate negative sol of As2S3, the coagulating power of different cations has been found to decrease in the order as :

Similarly, to coagulate a positive sol., such as Fe(OH)3, the coagulating power of different anions has been found to decrease in the order :

(iii) Dialysis. The method used to separate the impurities from the colloidal solution is called dialysis. Its principle is based upon the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment. The bag is suspended in fresh water. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution. Dialysis can be used for removing HCl from the ferric hydroxide sol. The method is shown in Fig.

The ordinary process of dialysis is slow. To increase the process of purification, the dialysis is carried out by applying electric field. This process is called electrodialysis.

(iv) Tyndall effect. When a strong beam of light is passed through a true solution placed in a beaker, in a dark room, the path of the light does not become visible as shown in Fig. However, if the light is passed through a sol, placed in the same room, the path of the light becomes visible when viewed from a direction at right angle to that of the incident beam.
This phenomenon was studied for the first time by Tyndall and therefore it is called Tyndall effect. The cause to Tyndall effect is the scattering of light by colloidal particles i.e., these particles first absorb the incident light and then a part of it gets scattered by them. Since the intensity of the scattered light is at right angle to the plane of the incident light, the path becomes visible only when seen in that direction. The particles in true solution are too small in size to cause any scattering i.e., the Tyndall effect is not noticed in true solutions.

Question 29:

Give four uses of emulsions.

Answer:
  1. The concentration of the sulphide ore of a metal by froth floatation process involves the use of some oil such as pine oil. The oil forms emulsion with ore particles. When air is bubbled through the emulsion, it rises to the surface as foam and is skimmed off.
  2. The various pharmaceuticals and cosmetics available in liquid form such as cod-liver oil, B-complex, ointments etc. are emulsions of water-in-oil type. These are readily adsorbed in the intestines.
  3. The cleansing action of soap is based upon the formation of oil-in-water type emulsion.
  4. Milk which is an important constituent of our diet is an emulsion of fat in water.
Question 30:

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

Answer:

The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations.

Question 31:

Explain the terms with suitable examples :
(i) Alcosol (ii) Aerosol and (iii) Hydrosol.

Answer:

(i) Alcosol. It is a collodial sol of a solid in alcohol as dispersion medium.
(ii) Aerosol. It is a collodial dispersion of a liquid in a gas e.g., fog.
(iii) Hydrosol. It is a colloidal sol of a solid in water as dispersion medium.

Question 32:

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

Answer:

Colloid is not a substance but it is a state of substance. This statement means a particular substance may exist as a colloid under certain conditions and as a crystalloid under other conditions. For example, sodium chloride in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves like a crystalloid while the concentrated solution behaves as a colloid (called associated colloid). It is the size of the particle that decides the state in which the substance exists. If the size of the particles lies in the range of 10 nm to 100 nm, it is in the colloidal state.

Question 33:

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

Answer:

Clean surface facilitates the adsorption of desired species and help in surface studies.

Question 34:

Why is chemisorption referred to as activated adsorption ?

Answer:

Chemisorption involves formation of bond between gaseous atoms or molecules and the solid surface. This requires high activation energy. Thus, it is referred to as activated adsorption.

Question 35:

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

Answer:

At lower concentration, soap forms a normal electrolytic solution with water. After a certain concentration called critical micelle concentration, colloidal solution is formed.

Question 36:

What happens when gelatin is mixed with gold sol ?

Answer:

Gold sol is a lyophobic sol. Addition of gelatin stabilises the sol.

Question 37:

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

Answer:

Clouds are colloidal in nature and carry charge. Spray of silver iodide, (an electrolyte) on the clouds results in coagulation leading to artificial rain.

Question 38:

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

Answer:

Ice creams are emulsions which get stabilised by emulsifying agents such as gelatin.

Question 39:

What is collodion ?

Answer:

It is a 4% solution of nitrocellulose in a mixture of alcohol and ether.

Question 40:

Why do we add alum to purify water ?

Answer:

The colloidal impurities present in water get coagulated by added alum. This makes water potable.

Question 41:

What happens when electric field is applied to colloidal solution ?

Answer:

The charged colloidal particles start moving towards oppositely charged electrodes.

Question 42:

What causes brownian motion in colloidal dispersion ?

Answer:

Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium causes Brownian motion. This stabilises the sol.

Question 43:

A colloid is formed by adding FeCl3 in excess of hot water. What will happen if excess sodium chloride is added to this colloid?

Answer:

Positively charged sol of hydrated ferric oxide is formed and on adding excess of NaCl, negatively charged chloride ions coagulate the positively charged sol of hydrated ferric oxide.

Question 44:

How do emulsifying agents stabilise the emulsion ?

Answer:

The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

Question 45:

Why are some medicines more effective in the colloidal form ?

Answer:

Medicines are more effective in the colloidal form because of large surface area and are easily assimilated in this form.

Question 46:

Why does leather get hardened after tanning ?

Answer:

Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tanin which has negatively charged colloidal particles, it results in mutual coagulation.

Question 47:

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

Answer:

In Cottrell precipitator, charged smoke particles are allowed to pass through a chamber having a series of plates charged to very high potential (20,000 to 70,000 V) opposite to the smoke particles. Smoke particles lose their charge on the plates and get precipitated. Therefore, the gases coming out of the chimney become free of charged particles.

Question 48:

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

Answer:

On adding dispersion medium, emulsions can be diluted to any extent. The dispersed phase forms a separate layer if added in excess.

Question 49:

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

Answer:

The minimum quantity of an electrolyte required to cause coagulation of a sol is called its coagulating value. According to Hardy-Schulze rule, greater the charge on flocculating ion and smaller is the amount of electrolyte required for precipitation, higher is the coagulating power of coagulating ion.

Question 50:

Why does bleeding stop by rubbing moist alum ?

Answer:

Moist alum coagulates the blood and forms blood clot. This stops bleeding.

Question 51:

Why is Fe(OH)3 colloid positively charged, when prepared by adding FeCl3 to hot water ?

Answer:

The adsorption of positively charged Fe3+ ions by the sol of hydrated ferric oxide results in positively charged colloid.

Question 52:

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

Answer:

Physisorption involves weak van der Waals forces which decrease with rise in temperature. The chemisorption involves formation of chemical bond involving activation energy and is favoured by rise in temperature like any other chemical reaction.

Question 53:

What happens when dialysis is prolonged ?

Answer:

When dialysis is prolonged, the traces of electrolyte which stabilises the colloids are removed completely. This makes the colloid unstable and therefore coagulation takes place.

Question 54:

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

Answer:

The dye eosin is adsorbed on the surface of silver halide precipitate. Therefore, the precipitate is coloured.

Question 55:

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

Answer:

Activated charcoal acts as an adsorbent for various poisonous gases present in the coal mines.

Question 56:

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

Answer:

River water is muddy and contains charged particles of clay, sand and other materials. When the river water comes in contact with sea water, the electrolytes present in sea water coagulate the suspended colloidal particles which ultimately settle down at the point of contact. The point at which river and sea meet is the site for coagulation. The deposition of coagulated clay results in delta formation.

Question 57:

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

Answer:

The process of physisorption for example that of H2 on finely divided nickel, involves weak van der Waals’ forces. With increase in temperature, hydrogen molecules dissociate into hydrogen atoms which are held on the surface by chemisorption.

Question 58:

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

Answer:

After the reaction is complete between adsorbed reactants, the process of desorption is important to remove products and further create space for the other reactant molecules to approach the surface and react.

Question 59:

What is the role of diffusion in heterogeneous catalysis ?

Answer:

The gaseous molecules diffuse on to the surface of the solid catalyst and get adsorbed. After the required chemical changes the products diffuse away from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

Question 60:

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

Answer:

When gaseous molecules come in contact with the surface of a solid catalyst, the gaseous molecules are held with the surface of the catalyst by weak forces on the surface. This increases the concentration of reactants on the surface. Different molecules adsorbed side by side have better chance to react and form new molecules. This increases the rate of reaction. Moreover, adsorption is an exothermic process. The heat released in the process of adsorption is utilised in increasing the reactionrate.

Question 61:

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

Answer:

The optimum temperature range for enzymatic activity is 298-310 K. When the temperature is below or above this temperature range, enzymatic activity gets affected. Thus, during fever, when temperature rises above 310 K, the activity of enzymes may be affected.

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