Chemistry Unsolved Sample Paper Solutions CBSE Class 12
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.
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.
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
8. (d) (13/12)
Explanation: A + B + C → Products
r = –(dA/dt)
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.
The reaction can be represented as:
Explanation: Osmotic pressure is a colligative property and more the number of particles more the osmotic pressure.
π = iCRT
π = iCRT
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.
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:
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:
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,
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.
(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).
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.
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
30. (a) Nickel(II) chloride.
(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.
(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.
33. (a) (Oxidation at anode) :
(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.
The reaction with a higher value of E°, takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
(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.
As compound [A] on heating with aqueous ammonia gives compound [B], so [A] must be benzoic
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.
(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.
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.
(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.
(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.
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.
(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.
(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.
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