Answers to chemistry assignment III1.a) Explain the free radical mechanism of addition polymerization , taking vinyl chloride as an example.Answer :Addition polymerization are brought about by linking together simple unsaturated molecules such as alkenes (e.g. ethylene and propylene) or substituted alkenes (e.g. vinyl) For example. The conversion of ethylene into polyethylene

The mechanism of polymerization of alkenes involves free radicals. The mechanism consists of three steps.(i) Initiation step (ii) Propagation step and (iii) Termination step (i) Initiation step: For polymerization to take place, a small amount of initiator is necessary. Peroxides, oxygen, azo compounds etc are used as initiators. The initiator forms free radicals that initiate the polymerization reaction. The free radical furnished by the initiator attacks the double bond of a monomer forming monomer free radical, which leads to the second step propagation. Generation of initiator free radical.(C6H5COO)2 2 C6H5COO 2CO2 + 2C6H5( R)Benzyl Peroxide Benzyl Peroxide Free radical Benzyl Free radical R + CH2= CHCl R- CH2- CHCl Vinyl chloride Monomer free radical(ii) Propagation step: the free radical formed in the above step attacks another monomer molecule to form another free radical. This process continues resulting in the formation of long chain.R- CH2- CHCl + CH2= CHCl R- CH2- CHCl- CH2- CHCl Vinyl chloride Dimmer free radical R- CH2- CHCl- CH2- CHCl + n CH2 = CHCl R- [CH2- CHCl]n+1- CH2- CHCl Polymer free radical chain(iii) Termination step: The chain propagation is terminated when the free radicals mutually combine either by coupling or by disproportionation. Coupling: In Coupling a polymer molecule of long chain length is formed. R- [CH2- CHCl]n- CH2- CHCl + R- [CH2- CHCl]m- CH2- CHCl R- [CH2- CHCl]m+n- CH2- CHCl -CH2- CHCl R Polymer

1.b) Explain molecular weight of polymers.Answer:

Number average molecular weight (Mn):The number average molecular weight is defined as the total weight of polymer divided by the total number of molecules.Mn = Total weight, w Total number of particles, n= n1M1 + n2M2 + n1 + n2 + ------(Where n1, n2---etc are number of particles having mass M1, M2 etc.)Mn = niMi niWeight average molecular weight (Mw):

Molecular weight of each particle is multiplied by total weight of this species. These factors are added for all the size species and divided by the sum of weights of all individual sizes.Mw = W1 M1 + W2 M2 + ------ W1 + W2+ ------Mw = n1 M12 + n2 M22 + ------ (Since W1 = n1m1) n1 M1 + n2 M2 +-----Mw = niMi2 niMi

2) a) What is glass transition temperature? Discuss the factors affecting glass transition temperature.Answer :

Glass transition temperature (Tg)The glass transition is the temperature at which a polymer abruptly transforms form the glassy state (hard) to rubbery state (soft).Factors affecting Tg1. Presence of inherently rigid structure.The flexibility of the polymer chain is determined by the degree of freedom. Linear polymers with single bonds have high degree of freedom of rotation. The presence of inherently rigid structures in the polymer chain such as aromatics, cyclic structures or bulky groups on backbone hinder this rotation and increase in Tg.For example: Polyethylene Benzene ring in Polystyrene Tg = -110oC Tg = 100oC2. Presence of polar groupsPresence of polar groups in the molecular chain leads to strong cohesive forces which restrict the molecular Mobility. This leads to an increase in Tg.For example Nylon-66 Polypropylene Tg = 57oC Tg = -18oC Nylon-66 has strong intermolecular forces than polypropylene because of the presence of polar amide group.3. Presence of branching and cross-linking Increase in branching and cross-linking restrict the chain rotation and therefore Tg increases.4. Presence of plasticizersPlasticizers are low molecular weight compounds are added to plastics to increase their flexibility. They weaken the intermolecular forces between the polymer chains and decrease Tg. For example adding a small amount of plasticizers such as diisooctyl phthalate to PVC reduces its Tg from 80oC to below room temperature.5. Presence of Molecular weightAs molecular mass increases Tg also increases but Tg is not affected by molecular masses if degree of polymerization is above 250. 6. Presence of stereo regularity of the polymer. A Syndiotactic polymer has a higher Tg than Atactic polymer which in turn has higher Tg than its Isotactic stereoisomer.

2) b) Explain structure property relationship of polymers.Answer:On the basis of structure: Polymers can be classified into three types.(i). Linear polymer:If monomeric units are joined in the form of long straight chains are called linear polymers. e. g. Polyethylene, nylons and polyesters.

(ii). Branched chain polymer: A branched chain polymer is composed of a main chain with one or more substituent side chains or branches. E.g. low density polyethene (LDPE)

(iii).Cross linked polymer: Monomeric units are linked together to constitute a three dimensional network. They are hard, rigid, and brittle. e. g. Bakelite, epoxy, rubber etc.

3) Give the synthesis, properties and application of i) PlexiglassPolymethylmethacrylate (Plexi glass/Lucite): Manufacture: It is synthesized by polymerization of methylmethacrylate at 60-70oC in presence of H2O2 as initiator.Emulsion polymerization

Properties:1. It is clear, colorless and transparent polymer like glass. 2. It has high softening point of about 130-140C.3. It transmits 98% of the sunlight including ultravoilt. Applications:1. It is used for making Aircraft windows, TV screens, plastic signs, helmet visors, 2. It is mainly used for automotive application such as signal light, lenses, dials etc.3. Its sheets are used for signs, glazing skylights and decorative purposes.ii) TeflonPreparation, properties and applications of Polytetrafluroethylene (Teflon, PTFE)

Manufacture: It is synthesized by polymerization of perfluoroethylene or tetrafluoroethylene at moderate pressure and at 60oC in presence of ammonium persulphate as catalyst.

Properties:1. It is highly-crystalline solid2. It is very tough and strong due to the presence of strong C F bonds3. It has very high softening point (above 350C). 4. It has very high chemical resistance to almost all solutions ( except hot alkali and hot fluorine) Applications:1. It is used in cooking ware and other anti-stick applications (due to non-wet ability).2. It is used for making non sticking stopcocks for burettes, and non lubricating cryogenic bearings.3. It is used for making gaskets, valve parts, pumps, pipes etc. 4. It is used as insulating material for motors, generators, coils, transforms and capacitors.

iii) PolyurethanesPolyurethanes:Manufacture:Polyurethanes have the characteristic urethane NHCOO- linkages in their repeat units. Such a linkage is prepared by the reaction of di-isocyanate with glycol in presence of tertiary amine (acts as catalyst).

Properties:1. It exhibits excellent resistance to abrasion and solvents2. It exhibits high tensile strength. Advantages/Application:1. It is used as foams, coatings, films, adhesives and elastomers. 2. Resilient polyurethane fibers are used for foundation garments and swim suits. 3. Rigid polyurethane foams are used as heat insulating materials.4. It is used for making bristles, Textile fibres, thin sheets etc.

4) a) What are elastomers? Explain the synthesis and application of silicon rubber.Answer :Elastomers: An elastomer is a linear polymer which exhibits elasticity and other rubber like properties.Silicone rubber:Manufacture: It is prepared by the reaction between a dialkyl dichlorosilane and water. This produces a hydroxyl intermediate, which condenses to form a polymer called silicone rubber.

Properties: 1. It is highly inert and resistant extreme environments and temperatures from -55oC to 300oC. Applications:1. It is used for making gaskets, foot wear, sportswear, artificial heart valves, sealants, non-stick cookware coatings etc.4) b) Give synthesis, application,and properties of epoxy resin.Answer :Epoxy resin: It is prepared by the condensation of excess of epichlorohydrine with bis phenol-A at 60oc in presence of aqueous NaOH. Applications:1. Epoxy resins are mainly used in surface-coating materials which give outstanding toughness flexibility, adhesion and chemical resistance.2. It can be used in both moulding and laminating techniques to prepare glass fibre-reinforced articles with good mechanical strength, chemical resistance and electrical insulating properties. 3. It is used as skid resistant for highway surfacing.4. To impart crease-resistance and shrinkage control on cotton and rayon fabrics.5. It is used as stabilizers for vinyl resins.5) a) What are polymer composites . Give application and synthesis of Kevlar fibre.Answer : Polymer composites: Two or more distinct components which combine to form a new class of materials suitable for structural applications are referred to as composite materials.Polymer composites are generally made of two components namely (i) Matrix (ii) Fiber. The matrix is usually a thermoset material such as epoxy resin or a polyamide. Fibre is most often a glass, but sometimes it may be carbon fiber, Kevlar. Kevlar:Manufacture:It is prepared by poly condensation between aromatic dichloride and aromatic diamines. Properties: Kevlar is exceptionally stronger; it has high heat stability and flexibilityApplications: It is used extensively in the aerospace and air craft industries. It is also used for making tires, industrial belts, bullet proof vests, high strength cloths, car parts (such as tyres, brakes, clutch lining) etc.5) b) What are conducting polymers? Explain the conduction in polyaniline.Answer : Conducting polymer:An Organic polymer with a highly delocalized pi electron system having electrical conductance of the order of a conductor is called conducting polymers. Conducting polymers are generally produced by doping an oxidizing or a reducing agent into an organic polymer with conjugated back bone consisting of pi-electron system.Examples: Polyaccetylene polypyrrole, polythiophene, polyaniline, etc.Polyaniline: Structure of polyaniline: Applications: 1. It is used as electrode material for commercial rechargeable batteries. 2. As conductive tracks on printed circuit board.3. As film membrane for gas separation. 4. It can be used to produce smart windows that absorb sunlight and control solar energy.5. It is used to control electromagnetic radiation.

Mechanism of Polyaniline: Polyaniline is partially oxidized, first using a ammonium peroxy disulphate into a base form of aniline which contains alternating reduced and oxidized forms of aniline polymer backbone. This base form of aniline when treated with aqueous HCl (1M) undergoes protonation of imine nitrogen atom; creating current carrying charged sites (+ve) in the polymer backbone. These charges are compensated by anions (Cl-).

6) a) Discuss the types of impurities present in natural water.Answer :Impurities in water: Impurities present in natural waters in water may be broadly classified into four categories:1. Dissolved impurities: a) Inorganic salts: Cations: Ca2+, Mg2+, Na+, K+, Fe2+, Al3+, and sometimes traces of Zn2+ & Cu2+. Anions: Cl-, SO42-, NO3-, HCO3- and sometimes F- & NO2-b) Gases: CO2, O2, N2, oxides of N2 & sometimes NH3, H2Sc) Organic Salts 2. Suspended impurities: a) Organic: Oil globules, vegetable & animal matter b) Inorganic: Clay and sand3. Colloidal impurities: a) Organic: Organic waste products, humic acids, coloring matter, complex protein, amino acids.b) Inorganic: Finely divided Clay and silica, aluminium hydroxide, ferric hydroxide.4. Bacterial impurities: Bacteria, other microorganisms and other forms of animal and vegetable life.

6) b) Discuss the boiler troubles in scale and sludge formation.2. Scale and sludge formation:Salts formed in the boilers due to evaporation can be removed in the form of precipitation. If the precipitated matter is soft, loose and slimy, it is called sludge. If the precipitated matter is hard and adherent on the inner walls of the boiler, it is called scale. Sludge can easily be scrapped off with a wire brush. Sludges are formed by substances which have grater solubility in hot water than in cold water. Example. MgCO3, MgCl2, CaCl2, MgSO4 etc. Disadvantages of sludge formation:i) Sludges are poor conductor of heat, so they tend to waste a portion of heat generated.ii) If sludges are formed along with scales, then sludge gets entrapped in the scales and both get deposited as scales. iii) Excessive sludge formation settles in the regions of poor water circulation such as pipe connection, plug opening etc., thereby causing choking of the pipes. Disadvantages of scale formation:i) Wastage of fuel: Scales acts as insulator and hence reduced rate of heat transfer and evaporative capacity of the boiler. Thus scale formation will result in wastage of the fuel and reduction in boiler efficiency. The wastage of fuel depends upon the thickness and the nature of scale.ii) Lowering of boiler safety: Overheating is done in order to maintain a constant supply of steam, makes the boiler material softer and weaker and this causes distortion of boiler tube and makes the boiler unsafe to bear the pressure of steam. Rapid reaction between water and iron occurs at high temperature, causing additional thinning of the tube wall. 3 Fe + 4 H2O Fe3O4(s) + 4 H2(g)iii) Decrease in efficiency: Excessive scaling may cause clogging of tubes. Considerable quantities of sludges may also be entrapped in the scale which may reduce the water circulation and impair the efficiency of the boiler. iv) Danger of explosion: When thick scales crack, due to uneven expansion, the water suddenly in contact with over-heated iron plates, resulting in the formation of a large amount of steam. So suddenly high pressure is developed, which may cause even explosion of the boiler. 7a) Write a note on corrosion in boilersBoiler corrosion is destruction / deterioration of boiler tubes, plates, pipelines by chemical or electrochemical attack by its environment. Reason for boiler corrosion: i) Dissolved oxygen: Dissolved oxygen in water in presence of high temperature, attacks boiler material. 2 Fe + 2 H2O + O2 2 Fe (OH)2 2 Fe (OH)2 + O2 2 [Fe2O3. 2 H2O]Rustii) Dissolved CO2: H2O +CO2 H2CO3 which has slow corrosive effects on the boiler material. CO2 is also released inside the boiler of water contains bicarbonates Mg (HCO3)2 MgCO3 + H2O + CO2H2CO3 is responsible for pitting corrosion.iii) Acids from dissolved salts: water containing dissolved magnesium salts liberate acids on hydrolysis. MgCl2 + 2 H2O Mg (OH)2 + 2 HClThe liberated acid reacts with boiler producing HCl again and again.Fe + 2 HCl FeCl2 + H2Fe Cl2 + 2 H2O Fe(OH)2 + HCl , Presence of even a small amount of MgCl2 will cause corrosion of iron to a large extent.7b)Define COD. Explain method of determining CODChemical Oxygen Demand (COD): It is the amount of oxygen in milligrams required to oxidize organic and inorganic compounds present in one dm3 of waste water using strong oxidizing agent K2Cr2O7.Principle: COD determination is based on dichromate reflux method because of higher oxidizing ability of dichromate. The organic matter gets oxidized completely by potassium dichromate (K2Cr2O7) with silver sulphate as catalyst in the presence of concentrated H2SO4 to produce CO2 and H2O. The excess K2Cr2O7 is titrated with ferrous ammonium sulphate [Fe (NH4)2(SO4)2]. The dichromate consumed gives the oxygen (O2) required for oxidation of the organic matter. The chemical reactions involved in the method are: 2K2Cr2O7 + 8 H2SO4 2 K2 SO4 + 2Cr2(SO4)3 + 8 H2O + 3O2 C6H12O6 + 6O2 6CO2 + 6H2O Cr2O72-+ 6Fe2+ + 14H+ 6Fe3+ 2Cr3+ + 7H2O

Procedure:Back Titration: Pipette out 25 cm3 of the waste water sample & 10 cm3 of standard potassium dichromate solution (using pipette) into a conical flask. Add one test tube of 6N sulphuric acid (containing silver sulphate & mercuric sulphate) with constant shaking. Contents of the conical flask are Reflux for about 30 minutes on a water bath. The solution mixture is cooled to room temperature. Add 2-3 drops of ferroin indicator and titrate against standard Mohrs salt solution until the color changes from bluish green to reddish brown. The titre valve is noted (say V1 cm3)Blank Titration:Repeat the above procedure by only 10 cm3 of standard potassium dichromate without waste water. The titre valve is noted (say V2 cm3).Calculation: Normality of FAS solution= N1Amount of K2Cr2O7 reacted with organic matter in terms of FAS= (V2 V1) ml Volume of water sample taken for titration= 25 ml 1000 ml of 1N of FAS solution= 8 gms of oxygen1ml of 1N of FAS solution= 8 mg of oxygen (V2 V1) ml of N1 of FAS solution = (V2 V1)X N1 X 8 mg of oxygen 25ml of water sample = (V2 V1)X N1 X 8 mg of oxygen1000 ml of water sample= (V2 V1)X N1 X 8 X 1000 mg of oxygen 25COD of the water samples = C mg/liter of Oxygen

8a)Define BOD. How does it differ from COD? Explain the method of determination of BOD of the water sample.Biochemical Oxygen Demand (BOD): It is defined as the amount of oxygen required in milligrams by the microorganism to bring about oxidation of biological compounds present in 1 dm3 waste water over a period of five days at 20oC.COD (Chemical Oxygen Demand) is the amount of oxygen required to degenerate all poution in a chemical way (by adding oxidising agents and heating). In general with chemical destruction you can remove more polution than with the biological way.As BOD is only a measurement of consumed oxygen by aquatic microorganisms to decompose or oxidize organic matter and COD refers the requirement of dissolved oxygen for the oxidation of organic and inorganic constituents both. Hence COD must be greater than BODPrinciple: The Biochemical Oxygen Demand (BOD) test measures the oxygen utilized for the biochemical degradation of organic matter and oxidation of inorganic materials during a specified incubation period. Temperature effects are held constant by performing a test at fixed temperature. The methodology of BOD test is to compute a difference between initial and final Do of the samples incubation. DO is estimate by iodometric titration.Procedure:A definite volume of sewage sample is diluted to known volume with water which contains nutrients (1ml of phosphate buffer pH = 7.2 (K2HPO4, KH2PO4 , Na2HPO4 and NH4Cl), 1 ml of MgSO4,1 ml of CaCl2, and 1 ml of FeCl3 for bacterial growth and sufficient free oxygen. Equal volumes of diluted sample are filled into two BOD bottles (D1 and D5).Blank titration: Do content of D1bottle is determined immediately. 2ml of MnSO4, 3 ml of alkaline KI are added into the D1 bottle and mix well (for about 10-15minutes) and allowed to stand for 2 minutes. 1ml of con H2SO4 is then added and shaken well to dissolve the ppt. A known volume of this solution is against standard sodium thiosulphate using starch as indicator till the disappearance of blue color. The titre valve is noted (say V1 ml)

Sample Titration: The D5 bottle is incubated for 5 days at 20oC. After 5 days unconsumed DO is measured as described above. N sample (D1) =N sample (D1) = NNa2S2O3 X Vna2S2O3 Vsample(D1)Amount of Oxygen available in sample (D1 bottle) = Nsample (D1 X eq.wt. of oxygen = N2 X 8 gmsAmount of Oxygen available in sample (D1 bottle) = N sample (D1 X 8000 mg of oxygenAmount of Oxygen available in sample (D1 bottle) = D1 mg of oxygen Amount of Oxygen available in Incubated bottle (D5 bottle) = D5 mg of oxygen BOD of water sample = (D2 D5) X Volume of water sample after dilution Volume of water sample before dilution8b)Explain the determination of dissolved oxygen by winkler method. Give the reactions involved.Principle: DO in water is determined by Winkers method. This method is based on the indirect iodometric titration. Manganese sulphate reacts with alkaline solution to give manganese hydroxide. Manganese hydroxide reacts with DO to give basic manganese hydroxide and precipitated. Precipitation is dissolved by addition of con H2SO4 and nascent oxygen is liberated. Nascent oxygen oxides potassium iodide to iodine. The liberated iodine is titrated against standard Na2S2O3using starch as indicator.

ROCEDURE: Determination of Dissolved Oxygen: Pipette out 25 cm3 of the waste water sample into a 250 ml conical flask. Add 2ml of MnSO4, 3 ml of alkaline KI and mix well (for about 10-15minutes) and allowed to stand for 2 minutes. 1ml of con H2SO4 is then added and shaken well to dissolve the ppt. A known volume of this solution is against standard sodium thiosulphate using starch as indicator till the disappearance of blue color. The titre valve is noted (say V1 ml).

CALCULATION: Normality of Na2S2O3 solution= N1Volume of Na2S2O3 solution rundown= V1 ml Volume of water sample taken for titration= 25 ml 1000 ml of 1N of Na2S2O3 solution= 8 gms of oxygenV1 ml of N1 of Na2S2O3 solution= V1X N1 X 8 gms of oxygen 100025ml of water sample = V1X N1 X 8 gms of oxygen 1000106 ml of water sample= V1X N1 X 8 X 106 gms of oxygen 1000 X 25Do of the water samples = D PPM of Oxygen

9a)Explain secondary treatment on sewageSecondary treatment (biological treatment/ activated sludge process): Activated sludge containg microorganisms are sprayed over the waste water (which is left after primary reatment). The microorganisms present in the sludge form a thin layer & thrive on the organic wastes in the sewage. Air is passed vigorously into the waste water in order to bring good contact between the organic wastes & bacteria in presence of air & sunlight. Under these conditions, aerobic oxidation of organic matter occurs. The sludge formed is removed by settling or filtration. A part of the sludge is reused & the rest is used as fertilizer. The residual water is chlorinated to remove bacteria & finally discharged into running water or used for watering plants.9b)discuss the purification of water by ion exchange method.Ion exchange or de-ionization or de-mineralization: Ion exchange resins are insoluble, cross-linked, long chain organic polymers with a microporous structure and functional groups are capable of exchanging their ions with cations (Acidic functional groups COOH, SO3H ) and anions( basic functional groups NH2, = NH). There are two types of Ion exchange resins. i) Cation exchange resins (RH+) are mainly styrene divenyl benzene co-polymers, which on sulphonation or carboxylation, become capable to exchange their hydrogen ions with the cations in the water.

i) Anion- exchange resins (ROH-) are mainly styrene divenyl benzene co-polymers, which contains amino or quaternary ammonium or quaternary phosphonium or tertiary sulphonim groups as an integral part of the resin matrix. On treatment with dil NaOH solution become capable to exchange their hydroxyl ions with the anions in the water. Process: The hard water is passed first through cation exchange column, which removes all the cations like Ca2+, Mg2+ etc. from it and equivalent amount of H+ ions are released to water. 2 RH++ Ca2+ R2Ca2+ + 2H+RH++ Na+ RNa+ + H+Then hard water is passed through anion exchange column, which removes all the anions like SO42-, Cl- etc. from it and equivalent amount of OH- ions are released to water. R OH-+ Cl- RCl- + OH-2R OH-+ SO42- R2 SO42- + 2 OH-Now H+ and OH- ions are combined to produce water molecule. H++ OH- H2OWater coming out from the exchanger is free from cations as well as anions and is known as deionized or demineralized water.

10)Define desalination of water. Explain the electrodyalysis of desalination of water.Desalination:The process of removal of dissolved salts from sea water and making it potable is called desalination.Electrodialysis:Principle: Passage of an electric current through a solution of salt results in migration of cations towards the cathode & anions towards the anode. The use of semi permeable cation or anion exchange membrane in an electrolytic vessel permits the passage of only cations or anions respectively in the solution. An electro dialyzer consists of a chamber carrying a series of compartments fitted with closely spaced alternate cation & anion (A) exchange semi permeable membranes between the electrodes. An electro dialyzer unit will have 200 to 1000 compartments.The feed water is taken in the dialyzer & the electrodes are connected to a source of an electric current.

The anions pass through the anion permeable membrane towards the anode. However, these ions do not pass through the next membrane which is permeable only to cations. Similarly the cations moving in the other direction will pass through the cation exchange membrane but not the next. These anions & cations collect in the alternate chambers; the water in these is enriched with salt while that in the other compartments is desalinated. The enriched & desalinated waters are withdrawn separately. The water containing less salt recycled further to reduce the salt content.

11)What are nano materials? how are they classified? Explain with examples.Nano materials:One nanometer is defined as one thousand-millionth of a meter (10-9 m).Size dependent properties of nano materials: 1. Surface area: physical and chemical properties of a material depend on its surface area. Surface area is increased on moving from bulk to nano scale. Nanomaterials have a significant proportion of atoms existing at surface. Properties like catalytic activity, gas adsorption and chemical reactivity depends on surface area. For example: Finely divided nickel acts as an effective catalyst in hydrogenation of oil than bulk Ni. 2. Electrical properties: The electronic bands in bulk material are continuous due to overlapping of orbitals of atoms. But in the nano size materials, very few atoms or molecules are present so the electronic bands become discrete. Hence, metals which are good conductors in bulk become semiconductors and insulators, at nano level. 3. Optical properties: The nanoparticles of different size, can scatter radiation of different wavelength. For example. Colors of few colloidal solutions. Nano particles of metals exhibit unique optical property called as surface Plasmon resonance. When light falls on the surface of metal, electrons starts oscillating back and forth in a synchronized way in a small space and the effect is called as Plasmon resonance.11b)Describe the solgel method of synthesis.1. Sol-gel process: Sol-gel process has been used in the synthesis of mono-dispersed nanoparticles of metal oxides and temperature sensitive organic-inorganic hybrid materials. It consists of following steps: i) Preparation of sol ii) Conversion of sol to gel iii) Aging of a gel iv) Removal of solvent v) Heat treatment i) Preparation of sol: Sol is prepared by dispersing metal oxide or metal alkoxide (precursors) in a solvent. ii) Conversion of sol to gel: Sol is then converted into a gel by hydrolysis and condensation of precursors. Hydrolysis and condensation reactions are initiated by addition of an acid or a base iii) Aging of a gel catalyst.Gel on aging condenses to nanoscale clusters of metal hydroxides. iv) Removal of solvent: The encapsulated liquid can be removed from gel by evaporative drying. reactions are v) Heat treatment: The sample is then calcined to obtain nanoparticles( 1 to 100 nm).M(OC2H5)4 + x H2O M(OC2H5)4-x (OH)x + xC2 H5OH

M(OC2H5)4-x (OH)x + M(OC2H5)4-x (OH)x (OC2H5)4-x (OH)x-1 M-O-M (OH)x-1 (OC2H5)4-x + H2O

12)Discuss the synthesis of nanomaterials i) chemical vapor condensation ii )hydrothermal process4. Chemical Vapour Condensation (CVC):The evaporative source used in GPC (Gas Condensation Processing) is replaced by a hot wall reactor in the Chemical Vapour Condensation. In this method a metal organic precursors are converted into vapors in a reduced pressure atmosphere and passed into a heated furnace by helium gas, where they are later decomposed. The residence time of the precursor in the reactor determines films or particles are formed or both. By adjusting the residence time of the precursor molecules, gas flow rate and the pressure difference between the precursor delivery system & the main chamber, then the temperature of the hot wall reactor results in the fertile production of nanosized particles of metals and ceramics instead of thin films as in CVD (Chemical Vapour deposition) processing.