Oswal 36 Sample Question Papers CBSE Class 12 Chemistry Solutions


1. (a) A—T, G—C

Explanation: The structure of DNA molecule is a double helix structure. In this structure double helix are made up of polynucleotide chains which are held together by H-bonds. In these helixes the adenine (A) base is linked with thymine (T) by two H-bonds and guanine (G) is linked with cytosine (C) by three H-bonds as A ≡ T, and G ≡ C.

2. (c) 1.75 ×10–5 mol L–1

Explaination:        $$\land^\circ_{M(ch_{3}C00H)}=^{\land^\circ_{M_{(H^+)}}}+^{\land^\circ_{(CH{3}CH00{-)}}}$$

$$=350+50=400\space S\space cm^2\space mol^{-1}$$




$$=0.007×(5×10^{-2})^2=1.75×10^{-5}\space mol\space L^{-1}$$

3. (d) It provides surface for redox reaction.

Explanation: An inert electrode is a cell which provides surface for either oxidation or for reduction reaction by conduction of electrons through its surface but does not participate in the cell reaction.

4. (a) COCl3. 3NH3
Explanation: CoCl3.3NH3 complex: In this molecule, three chloride ions satisfy primary and also secondary valency. At room temperature, silver nitrate does not precipitate Cl. Hence, the complex compound behaves as a neutral non-conducting molecule. It may be written as [CoCl3(NH3)3].
5. (d) Mn2+


Electronic configuration of Mn2+ = [Ar]3d5 and Number of unpaired electrons = 5
Electronic configuration of Fe2+ = [Ar]3d6 and Number of unpaired electrons = 4
Electronic configuration of Ti2+ = [Ar]3d2 and Number of unpaired electrons = 2
Electronic configuration of Cr2+ = [Ar]3d4 and Number of unpaired electrons = 4
Magnetic moment = (√n(n + 2)) BM, where n is the number of unpaired electrons
Therefore, the maximum number of unpaired electrons are in Mn+2.

6. (a) Equal to second

Explanation: 12 g of urea =(12/60) mol=0.2 mole

68.4 g of sucrose =(684/342)=0.2 mole

As mole fraction of both the solutes are same, the lowering of vapour pressures will be equal to second.

7. (c) A mixture of mono-, di-, tri- and tetra-halogen derivatives

Explanation: When halogen reacts with an alkane in the presence of sunlight or heat leads to the formation of a haloalkane (alkyl halide). Depending on the proportion of the two reactants that are used, various products of different amount are produced. For example, in the case of methane (CH4), a large excess of the hydrocarbon favours formation of methyl chloride as the primary product; whereas, an excess of chlorine favours formation of chloroform (CHCl3) and carbon tetrachloride (CCl4). Similarly, in general a mixture of mono-, di-, tri- and tetra-halogen derivatives is formed.

8. (d) (13/12)

Explanation: A + B + C → Products

r = –(dA/dt)

= k[A]1/2 [B]1/3 [C]1/4

So order is,  ((1/2)+(1/3)+(1/4)=(13/12))

9. (c) Number of particles present in the solution

Explanation: Those physical properties of a solution which depend upon the number of particles in a given volume of the solution or the mole fraction of the solute are called colligative properties. They do not depend on the nature of the solute. The following four properties are colligative properties:

1. Lowering of vapour pressure of the solvent.

2. Elevation in boiling point of the solvent.

3. Depression in freezing point of solvent.

4. Osmotic pressure.

10. (b) N2
Explanation: The gas evolved when methylamine reacts with nitrous acid is nitrogen gas, N2 gas. The reaction can be represented as:

The reaction can be represented as:

diagram 46
12. (b) BaCl2 > NaCl > Glucose

Explanation: Osmotic pressure is a colligative property and more the number of particles more the osmotic pressure.

π = iCRT

Hence, BaCl2 > NaCl > Glucose since the concentration is same and i for BaCl2 is 3, NaCl is 2 and glucose is 1.
13. (b) Vitamin B12
Explanation: Vitamin B12 contains a metal atom, it is also called cyanocobalamin and contains cobalt (Co) atom.

π = iCRT

Hence, BaCl2 > NaCl > Glucose since the concentration is same and i for BaCl2 is 3, NaCl is 2 and glucose is 1.

14. (a) 4.008%

Explanation: Degree of dissociation:

15. (c) A is true, but R is false.

Explanation: The polypeptide chain in globular protein is folded around itself, giving rise to a spherical structure. These are usually soluble in water and all enzymes are globular protein. Thus, assertion is correct statement but reason is wrong statement.

16. (a) Both A and R are true and R is the correct explanation of A.

Explanation: The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series. This is because in first transition series, electrons are filled only in d orbitals but in second and third transition series, electrons are filled in d-orbitals and f-orbitals. The screening effect of f-orbital is lower than that of d-orbital. So, nuclear charge experienced by valence electrons in second and third transition series is more than that of the first transition series. Therefore, force of attraction between nucleus and electrons is more. +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements. Thus, both assertion and reason are correct statements but reason is not the correct explanation of assertion

17. (c) A is true, but R is false.

Explanation: 2-Deoxyribose C5H10O4 is a carbohydrate. It is a monosaccharide. It is a deoxy sugar, derived from the sugar ribose by loss of an oxygen atom. Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds that can be hydrolysed to polyhydroxy aldehydes or polyhydroxy ketones. Some carbohydrates like Rhamnose do not follow the formula Cx(H2O)y. Thus, assertion is correct statement but reason is wrong statement.

18. (a) Both A and R are true and R is the correct explanation of A.

Explanation: At equilibrium, the component which has more volatility is always in high concentration. The higher the vapour pressure of a liquid at a given temperature, the higher the volatility and the lower normal boiling point of the liquid. The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of the component. Thus, both assertion and reason are correct statements and reason is the correct explanation of assertion.


19. (a) Since the reactant is 25% decomposed in 40.5 minutes, the remaining reactant is 75%, i.e.,

at = 40.5 min., (a – x) = 0.75a

We know that,

20. Mass of the solution = 100 g

Density, ρ = Mass/ Volume

Since, Volume of the solution = Mass/Density


(a) Mole fraction is a dimensionless quantity as it is a ratio of moles of solute to the total moles of solution.

(b) Henry's law holds goods when;

1. Pressure of gas is low and temperature is high.

2. The gas is reactive with the solvent.



For weak electrolyte like acetic acid:

22. (a) The general electronic configuration of transition metal is (n –1)d1–10ns1–2.
(b) The second ionisation enthalpy of Cu is high as compared to the second ionisation enthalpy of Zn because the electronic configuration of Cu is 3d104s1. So, second electron needs to be removed completely from filled d-orbital which is very difficult. On the other hand, the electronic configuration of zinc, is 3d104s2. Therefore upto second ionisation enthalpy the removal of electron from 4s electron is easy whereas removal of third electron requires high energy as Zn2+ attains 3d10 electronic configuration which is fully filled and hence is very stable.

23. Lactose is composed of β-D galactose and β-D glucose. Therefore, on hydrolysis it gives β-D galactose and β-D glucose. The chemical equation can be represented as:

diagram 53

24. The -OH group attached to the benzene ring in phenol activates, it towards electrophilic substitution.
It directs the incoming group to ortho and para positions in the ring as these positions become electron rich due to electronic effect (also mesomeric effect) caused by -OH group. This factor is not present in benzene. Therefore, out of benzene and phenol, phenol is more easily nitrated because of the presence of -OH group in phenol which increases the electron density at ortho and para position.

25. The reactions of arene with iodine is carried out in the presence of an oxidising agent. This is because reaction of arene with iodine are reversible and require the presence of an oxidising agent to oxidise the HI formed during the reaction and shift the reaction towards forward direction. For example,

Halogenation Reaction:



26. (a) Due to equality in size of Zr and Hf, Nb and Ta, Mo and UI etc., the two elements of pair have the same properties.

This is because, There is a close similarity in physical and chemical properties of the 4d and 5d series of the transition elements much more than expected on the basis of usual family relationship. This is because 5d and 4d series elements have virtually the same atomic radii and ionic radii due to lanthanide contraction.

(b) Transition elements show variable oxidation states because their valence electrons are in two different sets of orbitals, that is (n-1)d and ns. The energy difference between these orbitals is very less, so both the energy levels can be used for bond formation. Thus, transition elements have variable oxidation states. For example, Mn (Z-25) has the highest number of unpaired electrons in the d-subshell, and it shows a high oxidation state (+7). Scandium (Sc) only exhibits +3 oxidation state in these series.

27. (a) Though both [NiCl4]2– and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2– is paramagnetic.
diagram 55
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d-orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.
(b) In the complex [Co(NH3)5Br]SO4, the sulphate ions are not directly bound to the metal atom and hence are ionisable so the sulphate ions in solution give precipitate with barium, chloride whereas in [Co(NH3)5SO4]Br, the sulphate ions are linked directly to the metal ion hence, are not ionisable.
(b) In the complex [Co(NH3)5Br]SO4, the sulphate ions are not directly bound to the metal atom and hence are ionisable so the sulphate ions in solution give precipitate with barium, chloride whereas in [Co(NH3)5SO4]Br, the sulphate ions are linked directly to the metal ion hence, are not ionisable.

(c) The important uses of coordination compounds are :

1. They are used in many quantitative and qualitative chemical analysis at laboratory and industrial level.

2. They are very important in biological systems as pigments (chlorophyll, a magnesium compound; haemoglobin, an iron compound) and enzymes (carboxypeptidase, carbonic anhydrase).

3. They are used as catalyst in many industrial processes, e.g. Wilkinson’s catalyst, [(Ph3P)3RhCl].

28. (i) – COOH group of benzoic acid is highly deactivating towards electrophilic attack on benzene ring, as it withdraws the electron density towards itself. Hence, Friedel-Craft’s reaction, involving an electrophilic attack of alkyl or acyl group in presence of AlCl3 (anhydrous) does not happen.

(ii) Higher the acidic strength, lower is the pKa value
As chloroacetic acid has electron withdrawing Cl group present at the α-carbon atom, it is a stronger acid and hence releases H+ ion easily compared to acetic acid.
Therefore acetic acid is weaker acid and have higher pKa value than chloroacetic acid.



For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane.

Further, the boiling point decreases with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of 1-chloropropane.

Hence, the given set of compounds can be arranged in the increasing order of their boiling points as:

Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane


As the starting compound is optically active it must have a chiral carbon with four different groups on it. As one functional group –Br (as the formula suggests), the structure is –(CH3)(C2H5)(C3H7)C(Br), which satisfies the above given condition. Now when it undergoes a nucleophilic substitution reaction (as aq. KOH is the reagent), it has 2 possibilities, SN 1 or SN 2 mechanism. As the starting compound has a tertiary carbon it can form a stable carbocation, it will take the SN 2 reaction pathway. Thus the reaction mechanism can be shown as :
diagram 58

30. (a) Nickel(II) chloride.

(b) Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. Each Cl ion donates a pair of electrons. The compound is paramagnetic since, it contains two unpaired electrons.

(c) In tetrahedral complexes, one s and three p-orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally.


31. (a) Order of the reaction is 1, Molecularity of the reaction 2.

(b) Rate law equation of the given equation Rate = k[C12H22O11]

(c) Sucrose is dextrorotatory (+ 66·5°). On hydrolysis with dilute acid sucrose gives equimolar mixture of D-glucose and D-Fructose. Since the laevorotation of fructose (– 92·4°) is larger than the dextrorotation of glucose (+ 52·7°), the hydrolysis product has laevorotation. Hence, hydrolysis of sucrose is called inversion of sucrose.

diagram 59
32. (a) This reaction is SN 2 type.
(b) Due to the presence of acidic or active hydrogen in ethanol (C2H5OH), it readily reacts with Na to form sodium ethoxide and hydrogen.


33. (a) (Oxidation at anode) :

The reaction at anode with lower value of Ecell is preferred and therefore, water should get oxidised to give O2 but on account of over potential of oxygen, Cl gets oxidised preferably, liberating Cl2 gas.
2. Conductivity of CH3COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.

(b) Y is weak electrolyte. On dilution weak electrolytes undergo dissociation and thus, there is a steep increase in molar conductivity observed. Whereas, in case of strong electrolytes, as they are already dissociated completely, dilution does not affect the conductivity very much. So, in this case X is a strong electrolyte.

(c) (i) At cathode: The following reduction reactions complete to take place at the cathode.

Ag+(aq) + e → Ag(s); E° = 0.80 V
H+(aq) + e →(1/2)H2(g); E° = 0.00 V

The reaction with a higher value of E°, takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode : Since Pt electrodes are inert, the anode is not attacked by NO3 ions. Therefor OH or NO3 ions can be oxidised at the anode. But OH ions having a lower discharge potential and get preference and decompose to liberate O2.

(ii) At cathode: The following reduction reactions complete to take place at the cathode.


The reaction with a higher value of E°, takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

At anode: The following oxidation reactions are possible at the anode.

E° is preferred. But due to the over potential of oxygen, Cl gets oxidised at the anode to produce Cl2 gas.


Compound ‘C’ of molecular formula C7H7N is formed as a result of Hoffmann bromamide degradation, hence, [C] must be an amine and compound [B] must be an amide. Amine with

As compound [A] on heating with aqueous ammonia gives compound [B], so [A] must be benzoic

(b) (CH3)3N < C2H5NH2 < C2H5OH

Alcohol has the highest boiling point because the extent of H-bonding is maximum as oxygen is more electronegative than nitrogen of amines. Tertiary amines have the lowest boiling point because of absence of hydrogen bonding as it has no hydrogen bonded to nitrogen atom in the tertiary amines.

(c) (CH3)2NH and (CH3)3N can be distinguished by use of Hinsberg’s reagent i.e., benzene sulphonyl chloride. (CH3)2NH being a 2° amine forms N, N-Dimethylbenzene sulphonamide ion which is insoluble in aqueous ROH. While (CH3)3N is a 3° amine and it does not react with Hinsberg’s reagent.

(d) Primary, secondary and tertiary amines can be identified and distinguished by Hinsberg’s test. In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C6H5SO2Cl). The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified using Hinsberg’s reagent.

1. Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide which is soluble in alkali.

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.

2. Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.


There is no H-atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.

3. On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.


(a) pKb of aniline is more than that of methylamine: Higher the pKb value lower is the basic strength.

In aniline due to resonance, the electrons on the N-atom are delocalised over the benzene ring.
Therefore, the electrons on the N-atom are less available to donate.

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.

(b) Ethylamine is soluble in water whereas aniline is not: Ethylamine when added to water forms intermolecular H-bonds with water. Hence, it is soluble in water.

But aniline does not undergo H-bonding with water to a very large extent due to the presence of a bulky hydropholic —C6H5 group. Hence, aniline is insoluble in water.
(c) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide: Methylamine is more basic than water. Therefore, in water, methylamine produces OH ions from water.

(d) Although amino group is o, p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).


For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(e) Aniline does not undergo Friedel-Craft’s reaction: A Friedel-Craft’s reaction is carried out in the presence of AlCl3 but AlCl3 is a Lewis acid while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt.

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated.
Hence, aniline does not undergo the Friedel-Craft’s reaction.

35. (a) The complete the reaction is as follows:

(b) The IUPAC name of the given compound is 4-Chloro-3methylheptanal.

(c) Aldehydes and ketones have high boiling point as compared to hydrocarbons and ethers of comparable molecular mass. It is due to the weak molecular association in aldehyde and ketone arising out of the dipole-dipole interactions.

These dipole-dipole interactions are weaker than intermolecular H-bonding. The boiling points of aldehydes and ketones are much lower than those of corresponding alcohols and carboxylic acids which possess inter molecular hydrogen bonding.

(d) The given acids can be arranged in their decreasing order of acidity as follows:


(e) Ethyl benzene undergoes oxidation reaction to provide benzoic acid when it is heated with potassium permanganate in basic conditions.



Answer 4.

(i) After having his dinner, Mr. Thompson went to the little chamber where Maggie was lying. He found a pair of large bright eyes looking at him from the snowy bed. The looks were tender, grateful and pleading. This gave him extreme joy. Joe Thompson sat down and for me first time, examined the child carefully under the lamplight. The tender face was attractive and full of childish sweetness on which suffering had not been able to leave its marks. This strengthened Mr. Thompson’s sympathy for Maggie.

(ii) On the first day, after returning from his work, Joe Thompson encountered the girl’s childish face for the first time. He sat down beside her and taking her soft little hand confirmed her name. She affirmed in a trembling voice that her name was Maggie. Mr. Thompson asked about her sickness, then her treatment, and about the pain. The girl replied that she had some pain but now as she was in the soft bed, it felt good and comfortable. She was satisfied and grateful to Mr. Thompson.

(iii) At first Mrs. Thompson was against taking care of Maggie and insisted her husband to send her back to the poorhouse immediately. Thompson reminded her of the Bible and explained to her that it was a small thing for them to keep that poor motherless child for a single night. The voice was very strong but simultaneously there was moisture in his eyes. Mrs. Thompson did not answer but a soft feeling crept into her heart. She then spent the whole day with Maggie and at night she made an effort to be indifferent to Maggie in front of her husband. She kept silent on that theme and gave the child a toasted slice of bread which was softened with milk and butter added with a cup of tea. This showed that the chords of her heart were struck with sympathy for the child.

(iv) Earlier Mrs. Thompson was adamant for sending the poor child back to the poorhouse. But with the passage of time, the harshness of her behaviour converted into softness and her heart began to melt towards little Maggie. The tenderness innocence, patience, gratitude, nature and purity of the child moved her a lot and she asked her husband to keep her for one or two days more on the pretext of her weakness and helplessness before sending her back. Finally, she gave up her decision of sending Maggie to the poorhouse and accepted her wholeheatedly.

(v) The sick and helpless child brouhgt light and happiness to Thompson’s house. She was a blessing for them. For a long period of time, it had been dark, cold and miserable because Mrs. Thompson had no one to take care or love. That is why she became a sore, irritable and ill-tempered woman. Now the sweetness of that sick child who was also thirsty for getting someone’s love became honey to her soul as she carried her in her heart as well as arms. As for Joe, there was not a single man in the whole neighbourhood who drank as precious wine as the Maggie came as an angel in disguise in their house and filled its dreary chambers with love.

Answer 5.

(i) Seeing Owens disturbed, Luz Long approached him and tried to calm him down by giving suggestions. Suddenly all the tension seemed to flow away from his body as the truth what Long told him struck him. Full of confidence, he drew a line, a full foot in back of the board and proceeded to jump and qualified with almost a foot to spare. That night he went to Long’s room and thanked him. They sat and talked for two hours on every topic.

(ii) When Owens finished his jump he found Luz Long beside him congratulating him. He gave him a firm handshake which was far from any jealousy. He did not bother about the wrath of Hitler and congratulated Owens. He failed but helped a capable world record holder to set other records.

(iii) Luz Long broke his own past record but did not win. But Jesse Owens set the Olympic record of jumping 26 feet 5 – 5/16 inches. Luz Long came to his side congratulating him by shaking his hands hard. Hitler was watching all this and he was not a hundred yards away from them. He could not tolerate the defeat of his athlete and instead of congratulating the other athlete, he glared at both of them.

(iv) Coubertin is said to be the founder of the modern Olympic Games. He believed that the most important thing in the Olympic Games is not winning but taking part. The essential thing in life is not conquering but fighting well. Luz Long did not win but he presented the true example of this spirit.

(v) Jesse Owens recalls the incident in Olympic village with Luz Long where he bonded with him over their sport event. Their conversation with each other laid the foundation of their friendship, but what made it special was Luz’s reaction towards Jesse’s victory which wasn’t fake at all. Jesse had broken an Olympic record and while the whole stadium was glaring at him, Luz shook his hand proudly with a smile, which was worth more than all the gold medals and cups Jesse had.

CBSE 36 Sample Question Papers Science Stream (PCM)

All Subjects Combined for Class 12 Exam 2023

CBSE 36 Sample Question Papers Science Stream (PCB)

All Subjects Combined for Class 12 Exam 2023

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