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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