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12th Class Chemistry Sample Paper - 3

Chemistry (Theory)
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Time Allowed: 3 hours

Maximum Marks: 70

General Instructions:

(a) All questions are compulsory.

(b) Question numbers 1 to 5 are very short answer questions and carry 1 mark each.

(c) Question numbers. 6 to 10 are short answer questions and carry 2 marks each.

(d) Question numbers 11 to 22 are also short answer questions and carry 3 marks each.

(e) Question number 23 carries 4 marks.

(f) Question numbers. 24 to 26 are long answer questions and carry 5 marks each

1. Define Kraft temperature.

Ans. The formation of micelles takes place only above a particular temperature called Kraft temperature

2. Name the type of semiconductor obtained when silicon is doped with boron.

Ans. p-type semiconductor

3. Write one similarity between physisorption and chemisorption.

Ans. Both increase with increase in surface area.

4. Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.

Ans.MnO4-

5. What type of colloid is formed when a liquid is dispersed in a solid? Give an example.

Ans. Gel, e.g., cheese, jellies.

6. Account for the following

i. XeF2 IS linear molecule without a bend.

ii. The electron gain enthalpy with negative sign for fluorine is less than that of chlorine, still fluorine is a stronger oxidizing agent than chlorine

Ans. (i) XeF2 has 2 bond pairs and 3 lone pairs. The electron arrangement is trigonal bipyramidal.The shape is linear because the lone pairs prefer the equatorial position in order to minimise the repulsion among the electron pairs.

(ii) It is due to

a. Low bond dissociation enthalpy of F-F bond.

b. High hydration enthalpy of F- ion.


OR


Define the following properties

i. Colligative properties

ii. Molality

Ans.(i) Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute.

(ii)Molality (m) is the number of moles of solute present in 1kg of solvent and it is expressed as:

Molality = Moles of solute/mass of solvent in kg

7. Define the following terms.

a. Molarity

b. Ideal solution

Ans.

(a) Molarity (M) is the number of moles of solute present in 1L of solution.

(b) The solution which Obeys Raoult's law at every range of concentration is known as Ideal solution

8. Using IUPAC norms write the formulae for the following

i. Sodium dicyanidoaurate

ii. Tetraamminechloridoitrito-N-platinum(IV) sulphate

Ans i. Na[Au(CN)2 ]

ii. [Pt(NH3)4CI(NO2)]SO4

9. Write the Colligative property which is used to find the molecular mass of macromolecules.

Ans. Osmotic pressure is used to find the molecular mass of macromolecules.

10.i. Write the IUPAC name of the isomer of the following complex:

[Pt(NH3)2Cl2]

ii. Write the formula for the following:

Tetraamineaquachloridocobalt (III) nitrate.

Ans. i. Diamminedichloridoplatinum (II)

ii [Co(NH3)4(H2O)Cl]NO3)2

11. Write the main use of

i. Quaternary ammonium salts of long chain amines

ii. Esters of benzoic acid

iii. Aromatic diazonium salts.

Ans.

i. Quaternary ammonium salts of long chain amines are used in detergents.

ii. Esters of benzoic acid are used in perfumes.

iii. Diazonium salts are mainly used in the manufacture of dyes

12. Account for the following:

i. Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.

ii. pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.

Ans.

i. Because the carboxyl group is deactivating and the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group.

ii. Nitro group is an electron withdrawing group (-I effect) so it stabilises the carboxylate anion and strengthens the acid/Due to the presence of an electron withdrawing Nitro group (-I effect).

13. Define the following with an example of each:

i. Polysaccharides

ii. Denatured protein

iii. Essential amino acids

Ans.

i. Carbohydrates that give large number of monosaccharide units on hydrolysis/ large number of monosaccharides units joined together by glycosidic linkage Starch/ glycogen/ cellulose.

ii. Proteins that lose their biological activity/ proteins in which secondary and tertiary structures are destroyed Curdling of milk

iii. Amino acids which cannot be synthesised in the body. Valine / Leucine.

14. A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism.

15. The following data were obtained during the first order thermal decomposition of N2O5(g) at a constant volume:

2N2O5(g)--->2N2O4(g)+O2(g)

Calculate the rate constant

16.Following compounds are given to you:

2-Bromopentane, 2-Bromo-2-methylbutane,1-bromopentane

i. Write the compound which is most reactive towards SN2 reaction.

ii. Write the compound which is optically active

iii. Write the compound which is most reactive towards β-elimination reaction.

Ans. 1-bromopentane

i. 1-bromopentane, it is primary halide therefore undergoes SN2 reaction faster.

ii. 2-Bromopentane as carbon number two is symmetriv carbon

iii. 2-Bromo-2-methylbutane,because tertiary alkyl halides on dehydrogenation form most substituted alkene which is more stable.

17.i. Write the principle of method used for the refining of germination.

ii. Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth floatation process preferably?

iii. What is the significance of leaching in the extraction of aluminium?

Ans.

i. The impurities are more soluble in the melt than in the solid state of the metal.

ii. PbS

iii. Leaching is significant as it helps in removing impurities like SiO2 ,Fe2O3, and TiO2, etc.from the powdered bauxite ore by heating it with concentrated solution NaOH.

18. Write one difference in each of the following:

i. Lyophobic sol and Lyophilic sol

ii. Solution and colloid

iii. Homogenous catalysis and heterogeneous catalysis

19. i. Name the branched chain component of starch.

ii. Ribose in RNA and deoxyribose in DNA differ in the structure around which carbon atom?

iii. How many peptide linkages are present in a tripeptide?

Ans.

i. Amylopectin.

ii. C-2

iii. Two peptide linkages.

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

Ans. An emulsion is a colloidal dispersion in which both dispersed phase and dispersion medium are liquids, and the two liquids involved are otherwise immiscible. Types of emulsions are as follows:

i. Oil in water, in which oil is the dispersed phase and water is the dispersion medium. Example: Milk is as emulsion of liquid fat dispersed in water.

ii. Water in oil, in which water is the dispersed phase and oil is the dispersion medium.

Example: Cod liver oil is an emulsion of water in oil in which water is the dispersed phase and oil is the dispersion medium.

21. Account for the following:

i. Primary amines (R-NH2) have higher boiling point than tertiary amines (R3N).

ii. Aniline does not undergo Friedel -Crafts reaction.

iii. (CH3)2NH is more basic than (CH3)3N in an aqueous solution

Ans.

i. Due to maximum intermolecular hydrogen bonding in primary amines (i.e. due to the presence of more number of H atoms), primary amines have a high boiling point in comparison to tertiary amines.

ii. Aniline does not undergo Friedel–Crafts reaction due to the acid–base reaction between basic compounds. Aniline and Lewis acid/Protic acid are used in the Friedel-Crafts reaction.

iii. In (CH3)3N, there is maximum steric hindrance and least solvation,but in (CH3)2NH,solvation is more and steric hindrance is less, lesser than in (CH3)3N.Although +I effect is less since there are two methyl groups,di-methyl amine is still a stronger base than tri-methyl amine.

22. i. Which one of the following is a food preservative?

Equanil, Morphine, Sodium benzoate

ii. Why is bithional added to soap?

iii. Which class of drugs is used in sleeping pills?

Ans.

i. Sodium benzoate is used as a food preservative, whereas equanil is a tranquilliser and morphineis an analgesic.

ii. Bithional is an antiseptic added to soaps to reduce odours produced by bacterial decomposition of organic matter on the skin.

iii. Tranquillisers relieve stress and fatigue by inducing a sense of well-being, so they are used in making sleeping pills

23. On the occasion of World Health Day, Dr. Satpal organized a 'health camp' for the poor farmers living in a nearby village. After check-up, he was shocked to see that most of the farmers suffered from cancer due to regular exposure to pesticides and many wore diabetic. They distributed free medicines to them. Dr. Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care, financial assistance, setting up of super-specialty hospitals for treatment and prevention of the deadly disease in the affected villages all over India.

i. Write the values shown by

a. Dr. Satpal

b. NHRC

ii. What type of analgesics is chiefly used for the relief of pains of terminal cancer?

iii. Give an example of artificial sweetener that could have been recommended to diabetic patients.

Ans.

i.a. Dr Satpal distributed free medicines. He is shows concern for welfare of others.

b. NHRC take responsibility and did its duty properly.

ii. Aspirin

iii. Aspartame

24. a. Write the products formed when CH3CHO reacts with the following reagents:

i. HCN

ii. H2N –OH

iii. CH3CHO in the presence of dilute NaOH

b. Give simple chemical tests to distinguish between the following pairs of compounds:

i. Benzoic acid and Phenol

ii. Propanal and Propanone

Ans. a.i.Acetaldehyde reacts with of hydrogen cyanide to form hydroxypropanenitrile

ii. Ethanal reacts with hydroxylamine to form ethanal oxime.

CH3CHO + H2NOH --->CH3–C=NOH + H2O

Ethanal oxime

iii. Two molecules of ethanal condense in the presence of dilute NaOH to form 3-hydroxybutanal.

b. i. When treated with sodium bicarbonate, benzoic acid gives brisk effervescence due to the evolution of carbon dioxide gas. No reaction takes place when phenol is treated with NaHCO3.

C6H5COOH + NaHCO3→C6H5COONa + H2O + CO2

ii. Propanal gives a positive Tollens’test, while propanone does not give a positive Tollens’test.

CH3CH2CHO + 2[Ag(NH3)2] ++ 3OH−→ CH3CH2COO−+ 2Ag + 4NH3 + 2H2O


OR


(a)Account for the following:

i.Cl -CH2COOH is a stronger acid than CH3COOH.

ii.Carboxylic acids do not give reactions of carbonyl group.

(b) Write the chemical equations to illustrate the following name reactions:

i. Rosenmund reduction

ii. Cannizzaro's reaction

(c)Out of CH3CH2-CO -CH2-CH3and CH3CH2-CH2-CO–CH3which gives iodoform test?

Ans. (a)(i) Monochloroacetic acid is stronger than acetic acid. This is due to –Cl as a –I group.

(ii)Carboxylic acids do not give reactions of the carbonyl group because the lone pairs on oxygen attached to hydrogen in the –COOH group are involved in resonance which makes the carbon atom less electrophilic.

(b) i.Rosemund reaction:

ii. Cannizzaro’s reaction:

2HCHO + NaOH ---> CH3OH + HCOONa

(c) CH3CH2–CH2–CO –CH3

25. a. A reaction is second order in A and first order in B.

i. Write the differential rate equation.

ii. How is the rate affected on increasing the concentration of A three times?

iii. How is the rate affected when the concentrations of both A and B are doubled?

b. A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428 =0.1548).

Ans.

(a)(i)Differential rate equation: Rate = -d[R]/dt = K[A]2[B]

(ii)On increasing the concentration of A three times as 3A:

Rate = K[3A]2[B]

= 9K[A]2[B]   (i.e. New rate is 9 times the initial rate.)

= 9K[A]2[B]

= 9 (Rate)

(iii)On increasing the concentration of A and B two times as 2A and 2B:

= K[2A]2[2B]

= K(4*2)[A]2[B]

= 8K[A]2[B]

= 8 (Rate)

8 times the initial rate.


OR


a. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. Rate constant 'k' of a reaction varies with temperature 'T' according to the equation:

log K = log A - (Ea/2.303R)(1/T)

Where Ea is the activation energy. When a graph is plotted for log k Vs ( 1/T), a straight line with a slope of –4250 K is obtained. Calculate 'Ea' for the reaction. (R = 8.314 JK-1 mol-1)

Ans. For a first-order reaction,

Case 1: If 't' is the time required for 99% completion,then x = 99% of a

(a-x) = a/100

t=(2.303/K) log (a/(a-x))

t=(2.303/K) log ((a*100)/a)

t=(2.303/K) log 102

t= 2[2.303/K]

Case 2: If 't' is the time required for 90% of completion, then x = 90% of a

a-x = 10a/100 = a/10

t= (2.303/K) log (a/(a-x))

t=(2.303/K) log ((a*10)/a)

t= 2.303/K

Therefore, the time required for 99% completion of the first-order reaction is twice the time required for 90% completion.

b.log k = log A[(Ea/2.303R)](1/T)

Ea is activation energy.

The above equation is like y = mx + c, where if we plot yv/s x,we get a straight line with slope 'm' and intercept 'c'.

Slope = -(Ea/2.303R)

-(Ea/2.303R) = -4250K

Ea = 4250 * 2.303 * 8.314

Ea = 81375.3535 j mol-1

26.a. Give reasons for the following:

i. Bond enthalpy of F2 is lower than that of Cl2.

ii. PH3has lower boiling point than NH3.

b. Draw the structures of the following molecules:

i. BrF3

ii. (HPO3)3

iii. XeF4

Ans. (a).i. F atom is small in size. Due to this, the electron–electron repulsion between the lone pairs of F–F is large. Thus, the bond dissociation energy of F2 is lower than that of Cl2.

ii. PH3 has a lower boiling point than NH3 because the NH3 molecule possesses intermolecular hydrogen bonding which binds its molecules strongly, whereas PH3 has weaker Vander Waal's forces. Thus, PH3 has a lower boiling point than NH3.

(b) Structures of the following molecules:

i.BrF3, bent T-shape

ii.(HPO3)3, cyclic structure

iii.XeF4, square planar


OR


(a) Account for the following:

i. Helium is used in diving apparatus.

ii. Fluorine does not exhibit positive oxidation state.

iii. Oxygen shows catenation behavior less than sulphur.

(b) Draw the structures of the following molecules:

i. XeF2

ii. H2S2O

Ans.(a)i. Helium mixed with oxygen under pressure is given to sea divers for respiration. Air is not given to sea divers because nitrogen present in the air is soluble in the blood and will give a painful sensation called bends by bubbling out blood on moving from the high pressure in the deep sea to the low atmospheric pressure

ii. Fluorine being the most electronegative atom does not exhibit a positive oxidation state because its electrons are strongly attracted by the nuclear charge because of its small atomic size, and therefore, removal of an electron is not possible.

iii. Sulphur shows catenation behaviour more than that of oxygen because the sulphur atom is larger than the oxygen atom.

The O–O bond in oxygen experiences repulsion due to the lone pairs present on the oxygen atom, and therefore, the bond is weaker than the S–S bond.

(b)Structure of the following molecules:

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