CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi

Time allowed: 3 hours
Maximum Marks: 70

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I

Question 1.
‘Crystalline solids are anisotropic in nature’. What does this statement mean? [1]

Question 2.
Express the relation between conductivity and molar conductivity of a solution held in a cell. [1]
Answer:
The molar conductivity of a solution is related to the conductivity of that solution as:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q2

Question 3.
Define ‘Electrophoresis. [1]
Answer:
Electrophoresis is the phenomenon of movement of colloidal particles under the applied electric field.

Question 4.
What is the structure of XeF2 molecule? [1]
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q4

Question 5.
Write the IUPAC name of the following compound: (CH3)3CCH2Br [1]
Answer:
1-Bromo-2, 2-dimethylpropane

Question 6.
Draw the structure of 3-methylbutanal. [1]
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q6

Question 7.
Arrange the following compounds in increasing order of their solubility in water: [1]
C6H5NH2, (C2H5)2NH, C2H5NH2
Answer:
C6H5NH2 < (C2H5)2NH < C2H5NH2

Question 8.
Define Biodegradable polymers. [1]
Answer:
Biodegradable polymers are natural polymers that disintegrated themselves over a period of time by enzymatic hydrolysis, e.g., starch, cellulose, etc.

Question 9.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. [2]
Answer:
According to the electrochemical theory of rusting, the impure iron surface behaves like the small electrochemical cell in the presence of water containing dissolved oxygen or carbon dioxide. In this cell pure iron acts as an anode and impure iron, the surface acts as the cathode. Moisture having dissolved CO2 or O2 acts as an electrolyte. The reactions at cathode and anode are as follows:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q9

Question 10.
Determine the values of equilibrium constant (Kc) and ΔG0 and for the following reaction: [2]
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), E0= 1.05V (1F = 96500 C mol-1)
Answer:
According to the formula
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q10

Question 11.
Distinguish between ‘rate expression’ and ‘rate constant of a reaction. [2]
Answer:
The rate constant is the rate of reaction when the concentration of each reactant is taken as unity.
Rate expression expresses the rate of reaction in terms of molar concentrations of the reactants with each term raised to their power, which may or may not be same as the stoichiometric coefficient of that reactant in the balanced chemical equation.

Question 12.
Give the reason for: [2]
(i) The N-O bond in NO2 is shorter than the N-O bond in NO3
(ii) SF6 is kinetically an inert substance.
OR
State reasons for each of the following:
(i) All the P-Cl bonds in PCl5 molecules are not equivalent.
(ii) Sulphur has a greater tendency for catenation than oxygen.
Answer:
(ii) SF6 is kinetically inert due to high oxidizing power and electronegativity of fluorine atom which causes steric hindrance and it unable to further react with any other atom.
OR
(ii) Sulphur has a much greater tendency for catenation than oxygen because of its bigger size and low electronegativity due to which the S-S bond is stronger than O-O bond and there is more interelectronic repulsion in O-O than in S-S bond.

Question 13.
Assign reasons for the following: [2]
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibit a greater range of oxidation states than lanthanoids.
Answer:
(i) In aqueous solution, Cu+ undergoes disproportionation to form a more stable Cu2+ ion.
2Cu+(aq) → Cu2+(aq) + Cu(s)
The higher stability of Cu2+ ion in aqueous solution may be attributed to its greater negative ∆hydH than that of Cu+ ion. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu2+ ions.

(ii) Actinoids exhibit a greater range of oxidation states than lanthanoids. This is because there is less energy difference between 5f and 6d orbitals belonging to actinoids than the energy difference between 4f and 5d orbitals in case of lanthanoids.

Question 14.
Explain the following giving one example for each: [2]
(i) Reimer-Tieman reaction.
(ii) Friedel Craft’s acetylation of anisole.
Answer:
(i) When phenol is treated with chloroform in the presence of aqueous NaOH at 340 K followed by hydrolysis gives salicylaldehyde.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q14

(ii) When anisole is treated with acetyl chloride in the presence of anhydrous AlCl3, 2-methoxy acetophenone is formed.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q14.1

Question 15.
How would you obtain: [2]
(i) Picric acid (2, 4, 6-Trinitrophenol) from phenol.
(ii) 2-Methylpropene from 2-methyl propanol?
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q15

Question 16.
What is essentially the difference between the α-form of glucose and β-form of glucose? Explain. [2]
Answer:
In α-form of glucose, OH group lies at Cl towards the right but in β-form of glucose, it is towards left. This is because the OH group at Cl in glucose is chiral.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q16

Question 17.
Describe what you understand by primary structure and secondary structure of proteins.[2]
Answer:
Primary structure of proteins refers to the sequence in which amino acids are joined together by a peptide linkage. The sequence, of amino acids in primary structure, is very specific. Any change in the sequence of amino acids creates a different protein with different biological activity.

Secondary structure of proteins refers to the conformation which arises due to the coiling of the polypeptide chain due to intramolecular hydrogen bonding between carbonyl (=C=O) and -NH groups. Depending upon the size of -R group, there are two secondary structures of protein, i.e., α-Helix and β-Pleated sheet-like structure.

Question 18.
Mention two important uses of the following [2]

  1. Bakelite
  2. Nylon-6

Answer:

  1. Bakelite: It is used in the manufacture of electrical switches, protective coatings, barrels, etc.
  2. Nylon-6: It is used in making Fabrics, tyre cords, mountaineering ropes etc.

Question 19.
Silver crystallizes in a face-centred cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four comer atoms). [3]

Question 20.
Nitrogen pentoxide decomposes according to the equation: [3]
2N2O5(g) → 4NO2(g) + O2(g)
The first-order reaction was allowed to proceed at 40°C and the data below were collected:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q20
(i) Calculate the rate constant. Include units with your answer.
(ii) What will be the concentration of N2O5 after 100 minutes?
(iii) Calculate the initial rate of reaction.
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q20.1

Question 21.
Explain how the phenomenon of adsorption finds application in each of the following processes: [3]
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth Floatation Process
OR
Define each of the following terms:
(i) Micelles
(ii) Peptization
(iii) Desorption
Answer:
(i) In vacuum flask activated charcoal is placed between the walls of the flask so that any gas which enters into annular space either due to glass imperfection or diffusion through glass is adsorbed and create a vacuum.

(ii) If the catalysts and reactants are present in a different phase, the process of catalysis is called heterogeneous catalysis. For example, the manufacture of NH3 from N2 and H2 by Haber’s process using iron as a catalyst
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q21
In this process, the reactants are in the gaseous phase whereas catalyst is in a solid phase.

(iii) This method is used for removing gangue from sulphide ores. In this powdered ore is mixed with collectors (e.g., pine oil, fatty acids etc.) and froth stabilizers (e.g., cresols, aniline) which enhance non-wettability of the mineral particles and froth stabilisation respectively. As a result of which one comes with froth and gangue remains in the solution.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q21.1
OR
(i) When soaps and detergents are added to water, a cluster of the charged particle is formed by the aggregation of a variety of molecules. Thus formed is called micelle.

(ii) The process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte is called peptization.

(iii) The process of removal of the adsorbed substance from the surface of a solid or a liquid by heating or by reducing pressure is called desorption.

Question 22.
Describe the principle behind each of the following processes: [3]
(i) Vapour phase refining of a metal.
(ii) Electrolytic refining of a metal
(iii) Recovery of silver after silver ore was leached with NaCN.
Answer:
(i) Vapour Phase Refining: The impure metal is first converted to its unstable volatile compound which is evaporated and then decomposed by heating at the higher temperature to give pure metal, leaving behind the impurities.

(ii) In this method, impure metal is made anode and a thin sheet of pure metal is made cathode, and are put in a suitable electrolyte containing soluble salt of the same metal. On passing current the more basic metal remains in the solution and the legs basic one go to the anode and gets deposited as anode mud.

(iii) NaCN acts as a bleaching agent or oxidizing agent, thus oxidizes Ag to Ag+ which then combines with CN ions to form a respective soluble complex.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q22

Question 23.
Complete the following chemical equations: [3]
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q23
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q23.1

Question 24.
Write the name, stereochemistry and magnetic behaviour of the following: [3]
(At. Nos. Mn = 25, Co = 27, Ni = 28)
(i) K4[Mn(CN)6]
(ii) [CO(NH3)5Cl]Cl2
(iii) K2(Ni(CN)4]
Answer:
(i) Potassiumhexacyanomanganate(II)
Shape: Octahedral
Magnetic Behaviour: Paramagnetic (one unpaired electron)
Hybridization: d2sp3

(ii) Pentaamminechloridocobalt (III) chloride
Shape: Octahedral
Hybridization: d2sp3
Magnetic Behaviour: Diamagnetic (no impaired electrons)

(iii) Potassiumtetracyanonicklate(III)
Shape: Square planar
Hybridization: dsp2
Magnetic Behaviour: Diamagnetic (no impaired electrons)

Question 25.
Answer the following: [3]
(i) Haloalkanes easily dissolve in organic solvents, why?
(ii) What is known as a racemic mixture? Give an example.
(iii) Of the two bromo derivatives, C6H5CH(CH3)Br and C6H5CH(C6H5)Br, which one is more reactive in SN1 substitution reaction and why?
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attraction between haloalkanes and organic solvent molecules have the same strength as the one being broken in the separate haloalkanes and solvent molecules.

(ii) A racemic mixture is an equimolar mixture of two enantiomers and is hence optically inactive. e.g. (± butan-2-ol)

(iii) The carbocation intermediate derived from C6H5CH (C6H5)Br i.e. C6H5CHC6H5 is more stable as compared to the carbocation C6H5CHCH3 obtained from C6H5CH(CH3)Br because it is stabilized by two phenyl groups due to resonance.

Question 26.
(a) Explain why an alkylamine is more basic than ammonia? [3]
(b) How would you convert:
(i) Aniline to nitrobenzene
(ii) Aniline to iodobenzene
Answer:
(a) Alkyl groups are electron-donating groups and increase the electron density on nitrogen in alkylamine making them more basic than ammonia.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q26
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q26.1

Question 27.
Describe the following giving one example for each: [3]
(i) Detergents
(ii) Food preservatives
(iii) Antacids
Answer:
(i) Detergents are soluble salts of sodium-potassium sulphonic acids unlike soaps they are non-biodegradable but they can be conveniently used both with soft and hard water, e.g. Sodium alkylbenzene sulphonate.

(ii) Food preservatives are chemicals used to preserve food by preventing microbial growth, e.g. Sodium benzoate, Table salt, etc.

(iii) The substances which are taken to neutralize the excess acid and maintaining the pH to an appropriate level in the stomach are called antacids. There are two types of antacids systemic antacids, e.g., NaHCO3 and non-systemic e.g. Milk of magnesia.

Question 28.
(a) Differentiate between molality and molarity of a solution. How does a change in temperature influence their values? [5]
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol-1)
OR
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (Kf for water = 0.512 K kg mol-1, Molar mass of NaCl = 58.44 g)
Answer:
(a) Molarity is the number of moles of solute dissolved in 1 litre of solution. It is temperature-dependent.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q28
Molality is the number of moles of solute dissolved per 1 kg of the solvent. It is temperature independent.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q28.1
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.

(b) Since MgBr2 is an isotonic compound. Hence undergoes complete dissociation
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q28.2
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q28.3
OR
(a) Osmosis is the phenomenon of flow of solvent through a semi-permeable membrane from the region of higher concentration to the region of lower concentration.
The osmotic pressure of a solution is the excess pressure that must be applied to the solution to prevent the passage of solvent molecule through a semi-permeable membrane into the solution.
Yes, osmotic pressure is a colligative property as it depends only on the amount of solute present in the solution.

(b) Initial moles after dissociation
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q28.4

Question 29.
Give chemical tests to distinguish between: [5]
(i) Propanal and propanone
(ii) Benzaldehyde and acetophenone
(b) How would obtain:
(i) But-2-enal from ethane
(ii) Butanoic acid from butanol
(iii) Benzoic acid from ethylbenzene
OR
(a) Describe the following giving linked chemical equations
(i) Cannizzaro reaction
(ii) Decarboxylation
(b) Complete the following chemical equations:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29
Answer:
(a) (i) Propanal on treatment with Fehling’s solution gives a red ppt. of cuprous oxide while propanone does not respond to this test.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29.1

(ii) Acetophenone on treatment with I+2 NaOH undergoes iodoform test to give yellow ppt. of iodoform that benzaldehyde does not
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29.2
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29.3
OR
(a) (i) An aldehyde with no α-hydrogen atom undergoes self-reduction and oxidation in presence of cone, alkali to form alcohol and carboxylic acid salt.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29.4

(ii) Sodium acetate undergoes decarboxylation (removal of CO2) in the presence of soda lime to give hydrocarbon.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q29.5

Question 30.
(a) Explain the following: [5]
(i) NF3 is an exothermic compound whereas NCl3 is not.
(ii) F2 is most reactive of all the four common halogens.
(b) Complete the following chemical equations:
(i) C + H2SO4 (conc) →
(ii) P4 + NaOH+ H2O →
(iii) Cl2 + F2 →
OR
(a) Account for the following:
(i) The acidic strength decreases in the order HCl > H2S > PH3
(ii) A tendency to form pentahalides decreases down the group in group 15 of the periodic table.
(b) Complete the following chemical equations:
(i) P4 + SO2Cl2 →
(ii) XeF2 + H2O →
(iii) I2 + HNO3 (conc.) →
Answer:
(a) (ii) Fluorine is most reactive of all the four common halogens because of its low bond dissociation energy due to which it readily dissociates into atoms and reacts with other substances readily.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q30
(a) (i) Because of the decrease in electronegativity from chlorine to phosphorus, the dissociation enthalpy from HCl to HP increases, and their tendency to release H+ ion decreases and thus acidic strength decreases.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set I Q30.1

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in the previous set.

Question 1.
Which stoichiometric defect in crystals decreases the density of a solid? [1]

Question 3.
What is meant by shape-selective catalysis of reactions? [1]
Answer:
The reaction that depends for shape-selective catalysts uses zeolites as a catalyst for reaction on the shape and size of pores and of reactants and products.

Question 4.
Draw the structure of XeF4 molecule [1]
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q4

Question 9.
Explain what is meant by

  1. a peptide linkage
  2. a glycosidic linkage? [2]

Answer:

  1. Peptide linkage is present in proteins to bind together amino acids. The linkage involves the carboxyl group of one amino acid and amine group of another amino acids.
  2. Glycosidic linkage is the (-C-O-C-) linkage present between two molecules of a monosaccharide to form a disaccharide.

Question 10.
Name the bases present in RNA. Which one of these is not present in DNA. [2]
Answer:
Four bases present in RNA are Adenine, Guanine, Cytosine and Uracil. Uracil is not present in DNA.

Question 22.
Explain the role of each the following in the extraction of metals from their ores: [3]
(i) CO in the extraction of nickel.
(ii) Zinc in the extraction of silver.
(iii) Silica in the extraction of copper.
Answer:
(i) CO is used in the vapour phase refining of nickel.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q22

(ii) Zinc acts as a reducing agent which reduces the cyanide complex of silver into pure silver.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q22.1

(iii) Silica is used to remove impurities in the form of metal oxides as slag.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q22.2

Question 24.
For the complex [Fe(en)2Cl2]Cl, identify the following: [3]

  1. The oxidation number of iron.
  2. Hybrid orbitals and shape of the complex.
  3. Magnetic behaviour of the complex.
  4. The number of its geometrical isomers.
  5. Whether there may be optical isomer also.
  6. Name of the complex.

Answer:

  1. +3
  2. d2sp3 octahedral shape.
  3. paramagnetic
  4. 2 geometrical isomers, cis and trans.
  5. Only cis-isomers shows optical isomerism
  6. Dichloridobis (ethylenediamine) iron(III) chloride

Question 27.
Explain the following terms with one suitable example for each: [3]
(i) A sweetening agent for diabetic patients
(ii) Enzymes
(iii) Analgesics
Answer:
(i) Artificial Sweetening agents are chemicals that sweeten food. However, unlike natural sweeteners, they do not add calories to the body, not harmful to diabetic patients, e.g. Saccharin, aspartame.

(ii) Enzymes are biocatalysts which are structurally globular proteins. They are sensitive to a substrate, pH and temperature changes, e.g., Trypsin

(iii) Analgesics are chemical substances which reduces pain without causing impairment of consciousness, mental confusion, incoordination of paralysis or some other disturbance of nervous system, e.g., Aspirin, (non-narcotic analgesic) and Morphine (narcotic analgesic)

Question 28.
(a) State the following: [5]
(i) Henry’s law about partial pressure of a gas in a mixture.
(ii) Raoult’s law in its general form in reference to solutions.
(b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torrs at 25°C. Assuming the gene fragment is non-electrolyte, find its molar mass.
OR
(a) Difference between molarity and molality in a solution. What is the effect of temperature change on molarity and molality in a solution?

(b) What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform form a solution that has a boiling point of 68.04°C? The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant, Kb for chloroform is 3.63°C/m.
Answer:
(a) (i) Henry’s law states that the partial pressure of the gas in vapour phase is directly proportional to its mole fraction in the solution.

(ii) Raoult’s law states that for a solution with volatile components, the partial vapour pressure of each component present in the solution is proportional to its mole fraction in the solution.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q28
Substituting all the values in the given formula
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q28.1
OR
(a) Molarity is defined as the number of moles of solute present in 1 litre of solution, while molality is defined as the number of moles of solute present in 1 kg of solvent. Molality does not have any effect of change in temperature because mass does not changes with temperature, whereas molarity changes with temperature.
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set II Q28.2

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question 4.
Draw the structure of BrF3 molecule [1]
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q4

Question 8.
In nylon 6, 6 what does the designation 6, 6′ mean? [1]
Answer:
6, 6-refers to the number of carbon atoms in each of its monomer, adipic acid and hexamethylenediamine

Question 9.
What type of battery is lead storage battery? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery. [2]
Answer
The lead storage battery is a secondary cell which is rechargeable. During discharging, the electrode reaction occurs as follows:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q9

Question 10.
Two half-reactions of an electrochemical cell are given below:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q10
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q10.1
Construct the redox equation from the standard potential of the cell and predict it the reaction is reactant favoured or product favoured. [2]
Answer:
The redox reactions at anode and cathode can be represented as:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q10.2

Question 13.
Assign reasons for each of the following: [2]
(i) Transition metals generally form coloured compounds.
(ii) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements.
Answer:
(i) Due to the presence of unpaired electrons and d-d transition, the transition metals are generally coloured.

(ii) Manganese (Z = 25), has the maximum number of unpaired electrons. Thus, it shows oxidation states from +2 to +7 which is maximum in number as compared to other elements of transition series.

Question 18.
Name the sub-groups into which polymers are classified on the basis of the magnitude of intermolecular forces. [2]
Answer:

  1. Elastomers: They have weakest intermolecular forces of attraction.
  2. Fibres: They have strong intermolecular forces of attraction among its molecules.
  3. Thermoplastics Polymers: They are semifluid substances having low molecular weight.
  4. Thermosetting Polymers: They have inter-molecular forces intermediate between those of elastomers and fibres.

Question 19.
The density of lead is 11.35 g cm-3 and the metal crystallizes with fcc unit cell. Estimate the radius of lead atom. (At. mass of lead = 207 g mol-1 and NA = 6.02 × 1023 mol-1) [3]

Question 26.
Complete the following chemical equations: [3]
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q26
Answer:
CBSE Previous Year Question Papers Class 12 Chemistry 2011 Delhi Set III Q26.1

Question 27.
Answer the following questions: [3]
(i) Why do soaps not work in hard water?
(ii) What are the main constituents of Dettol?
(iii) How do antiseptics differ from disinfectants?
Answer:
(i) Hard water contains insoluble chloride of calcium and magnesium which forms insoluble ppt. (scum) with soap and thus cannot be rinsed off easily.

(ii) The main constituents of Dettol are chloroxylenol and α-terpene.

(iii) Antiseptics: These are chemical substances which either kill or prevent the growth of micro-organism but do not cause harm to the living tissues.
Disinfectants: These are chemical substances which kill the microbes. They are toxic in nature and thus causes harm to the tissues of the skin.

CBSE Previous Year Question Papers