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12th Class Chemistry Sample Paper - 2

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. Agl crystallises in cubic close packed ZnS structure. What fraction of tetrahedral sites are occupied by Ag+ ion?

Ans. Half of the voids of tetrahedral sites are occupied by Ag+ ion.

2. When dried fruits and vegetables are placed in water for some time, they slowly swell up. Give reason.

Ans. When dried fruits and vegetables are placed in water for some time, they slowly swell up as the process of osmosis takes place.

3. An element occurs in the BCC structure. How many atoms are present in its unit cell.

Ans. 2

4. What happens when blood cells are placed in pure water? Write two ions which are responsible for maintaining proper osmotic pressure inside and outside the living cells.

Ans. If blood cells are placed in pure water ,water molecules move into the blood cells and blood cells swell.Na+ and K+ ions are responsible for maintaining proper osmotic pressure inside and outside the cells.

5. Based on the type of dispersed phase, what type of colloids are micelles?

Ans. Associated colloids

6. Calculate the molality of ethanol solution in which the mole fraction of water is 0.88.

Ans. Mole fraction of water, χH2O=0.88

Mole fraction of ethanol, χC2H5OH = 1-0.88 =0.12

χC2H5OH = (n2)/(n1+n2)

n2= number of moles of ethanol.

n1= number of moles of water. Molality of ethanol means the number of moles of ethanol present in 1000 g of water.

n1= 1000/18 = 55.5 moles

Substituting the value of n1 in equation (1)

(n2)/55.5+n2)= 0.2

n2=7.57 moles

Molality of ethanol (C2H5OH) = 7.57 m

7. Write the structures and IUPAC names of the cross aldol condensation products only of ethanal and Propanal

Ans.

8. Name the crystal defect which reduces the density of an ionic solid? What type of ionic substances show this defect?

Ans. Schottky defect It is shown by ionic substances in which the cation and anion are of almost similar sizes. / ionic substances having high coordination number.

9. Draw the structure of the following compounds:

i. H2S2O7

ii. XeOF4

Ans.

10. Draw the molecular structures of the following:

i. Noble gas species which is isostructural with BrO3-

ii. Dibasic oxoacid of phosphorus

Ans.

11. State briefly the principles involved in the following operations in metallurgy. Give an example.

i. Hydraulic washing.

ii. Zone refining.

Ans.

i. Hydraulic washing: Principle involved: differences in gravities of the ore and the gangue particles e.g. oxide ores (haematite), native ores Au, Ag.

ii. Zone refining: Principle involved: the impurities are more soluble in the melt than in the solid state of the metal. e.g. germanium, silicon, boron, gallium and indium.

12. Give reasons for the following:

i. When 2g of benzoic acid is dissolved in 25 g of benzene, the experimentally determined molar mass is always greater than the true value.

ii. Mixture of ethanol and acetone shows positive deviation from Raoult’s Law.

iii. The preservation of fruits by adding concentrated sugar solution protects against bacterial action.

Ans.

i. Molecules of benzoic acid dimerise in benzene, the number of particles are reduced

ii. The intermolecular interactions between ethanol and acetone are weaker/ the escaping tendency of ethanol and acetone molecules increases on mixing / the vapour pressure increases.

iii. Due to osmosis, a bacterium on fruit loses water, shrivels and dies.

13. Explain what is observed when

i. Silver nitrate solution is added to potassium iodide solution.

ii. The size of the finest gold sol particles increases in the gold sol. Two oppositely charged sols are mixed in almost equal proportions.

Ans.

i. When silver nitrate solution is added to potassium iodide solution, a precipitate of silver iodide is formed which adsorbs iodide ions from the dispersion medium and a negatively charged colloidal solution is formed.

ii. As the size of the gold sol particles increases, the colour of the solution changes from red to purple, then blue and finally golden because the colour of colloidal solution depends on the wavelength of the light scattered by the dispersed particles and wavelength further depends on the size of the particles.

iii. When oppositely charged sols are mixed in almost equal proportions, neutralization of their charges occur and precipitation occurs.

14. Give reasons for the following

i. Use of aspartame as an artificial sweetener is limited to cold foods.

ii. Metal hydroxides are better alternatives than sodium hydrogencarbonate for treatment of acidity.

iii. Aspirin is used in prevention of heart attacks.

Ans.

i. It is unstable at cooking temperature.

ii. Excessive hydrogencarbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides being insoluble do not increase the pH above neutrality.

iii. Aspirin has anti blood clotting action.

15.

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.

16.i. Which of the following biomolecule is insoluble in water? Justify. Insulin, Haemoglobin, Keratin.

ii. Draw the Haworth structure for α-D-Glucopyranose.

iii. Write chemical reaction to show that glucose contains aldehyde as carbonyl group.

Ans.

i. Keratin is insoluble in water. It is a fibrous protein in which the polypeptide chains are held together by strong intermolecular forces, hence insoluble in water.

ii. α-D-Glucopyranose

iii. The sequence in the complimentary strand is

ATGCTTGA

17. Explain the following observations giving appropriate reasons:

a. Ozone is thermodynamically unstable with respect to oxygen.

b. The HEH bond angle of the hydrides of group 15 elements decrease as we move down the group.

c. Bleaching effect of chlorine is permanent.

Ans.

i. Decomposition of ozone into oxygen is an exothermic process (∆H = -ve) and results in an increase in entropy (∆S = +ve), resulting in large negative Gibbs energy change (∆G = -ve). Decomposition of ozone to oxygen is a spontaneous process.

ii. As we move down the group, the size of the central atom increases and the electronegativity decreases. Therefore the bond pair of electrons lies away from the central atom, the force of repulsion between the adjacent bond pairs decreases. So the bond angle decreases.

iii. The bleaching action of Cl2 is due to the oxidation of colored substance to colourless by nascent oxygen.

Cl2 + H2O → 2HCl + [O] Nascent oxygen.

18. a. Predict the number of unpaired electrons in the tetrahedral [MnBr4]2- ion.

b. Draw structures of geometrical isomers of [Co (NH3)4Cl2]+.

c. Write the formula for the following coordinate compound: Amminebromidochloridonitrito-N-platinate(II)

Ans.

i. Mn has the configuration 3d54s2. Hence the configuration of Mn2+ in [MnBr4]2- is 3d5.

Mn2=

Since it is tetrahedral in shape, the hybridization is sp3.

There are five unpaired electrons.

iii. [Pt(NH3)BrCl(NO2)]-

19. A metal complex having composition Cr (NH3)4Cl2Br has been isolated in two forms A and B. The form A reacts with AgNO3to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

i. Write the formulae of isomers A and B.

ii. State the hybridisation of chromium in each of them.

iii. Calculate the magnetic moment (spin only value) of the isomer A

Ans.

i. Isomer A: [Cr(NH3)4BrCl]Cl

Isomer B: [Cr(NH3)4Cl2] Br

ii. Hybridisation of Cr in isomer A and B is d2sp3.

iii. Number of unpaired electrons in Cr3+(3d3) is 3

Magnetic moment =√(n(n+2))= √(3(3+2)) =3.87BM

20. Complete the following reactions

Ans.

21. a. A colloidal sol is prepared by the given method in figure. What is the charge of AgI colloidal particles in the test tube? How is the sol formed, represented?

b. Explain how the phenomenon of adsorption finds application in Heterogeneous catalysis.

c. Which of the following electrolytes is the most effective for the coagulation of Fe(OH)3 sol which is a positively charged sol?

NaCl, Na2SO4, Na3PO4

Ans.

a. Negative charge is developed on the sol.

Sol is represented as AgI/ I-

b. Adsorption of reactants on the solid surface of the catalysts increases the rate of reaction.

c. Na3PO4 Hardy-Schulze rule

22. Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, calculate the density of Niobium. (Atomic mass = 93u)

23. John had gone with his mother to the doctor as he was down with fever. He then went to the chemist shop with his mother to purchase medicines prescribed by the doctor. There he observed a young man pleading with the chemist to give him medicines as he had nasal congestion. The chemist gave him cimetidine. John advised and also explained to the young man that he should only take the medicines prescribed by the doctor.

Answer the following questions:

i. Did the chemist give an appropriate medicine? Justify your answer.

ii. John’s action was appreciated by his mother. List any two reasons.

a. Write the mechanism of hydration of ethene to form ethanol.

Ans.

i. No the chemist did not give the appropriate medicine. Cimetidine is an antihistamine, but it is an antacid and not an antiallergic drug. Antacid and antiallergic drugs work on different receptors. Therefore cimetidine cannot be used to treat nasal congestion.

ii. Critical thinking

iii. Social responsibility

24. a. Account for the following:

i. Direct nitration of aniline yields significant amount of meta derivative.

ii. Primary aromatic amines cannot be prepared by Gabriel phthalimide synthesis.

b. Carry out the following conversions:

i. Ethanoic acid into methanamine.

ii. Aniline to p-Bromoaniline.

c. Arrange the following in increasing order of basic strength: Aniline, p-nitroaniline and p-toludine

Ans.

a. i. In strongly acidic medium, aniline is protonated to form the anilium ion which is meta directing.

ii. Aryl halides do not undergo nucleophilc substitution with the anion formed by phthalimide.

c. p-Nitroaniline < Aniline < p-Toludine

25. i. Graphically explain the effect of temperature on the rate constant of reaction? How can this temperature effect on rate constant be represented quantitatively?

ii. The decomposition of a hydrocarbon follows the equation

Calculate Ea

Ans.

i. For a chemical reaction with rise in temperature by 100, the rate constant is nearly doubled.

Increasing the temperature of the substance increases the fraction of molecules which collide with energies greater than Ea (activation energy)

As the temperature increases (T2), the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of reaction.

k = Ae-Ea/RT(Arrhenius equation)

ii. k = A e-Ea/RT

-Ea/RT = - 28000 K/T

Ea= 28000 K * 8.314/J/K/mol

Ea= 232.79 kj/mol

26. a. Account for the following observations:

i. SF4 is easily hydrolysed whereas SF6is not easily hydrolysed.

ii. Chlorine water is a powerful bleaching agent.

iii. Bi(V) is a stronger oxidising agent than Sb(V)

b. What happens when

i. White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2.

ii. (XeF6 undergoes partial hydrolysis.

(Give the chemical equations involved).

Ans. a.

i. S atom in SF4is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF6 is protected by six F aoms. Thus attack by water molecules cannot take place easily.

ii. Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.

CI2+ H2O ---> 2HCL+[O]

iii. Due to inert pair effect Bi (V) can accept a pair of electrons to form more stable Bi (III). (+3 oxidation state of Bi is more stable than its +5 oxidation state).

b. i. Phosphorus undergoes disproportionation reaction to form phosphine gas.

P4+3NaOH+3H2O ---> PH3+ 3NaH2PO2

ii. On partial hydrolysis, XeF6gives oxyfluoride XeOF4 and HF.

XeF6+H2O ---> XeOF4 + 2HF


OR


What inspired N.Bartlett for carrying out reaction between Xe and PtF6?

Arrange the following in the order of property indicated against each set:

i. F2, I2, Br2, Cl2 (increasing bond dissociation enthalpy)

ii. NH3, AsH3, SbH3, BiH3, PH3 (decreasing base strength

c. Complete the following equations

i. CI2+NaOH (cold and dilute) --->

ii. Fe3+ + SO2 + H2O --->

Ans.

a. N.Bartlett first prepared a red compound O2+PtF6-. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF6.

b.

i. I2< F2< Br2< Cl2

ii. NH3> PH3> AsH3> SbH3> BiH3

c.

i. 2NaOH+Cl2 ---> NaCl+NaOCI+H2O

ii. 2Fe3++SO2+2H2O --->2Fe2+ + SO2-4 + 4H+

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