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12th Class Chemistry Sample Paper - 1

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. Give one example of piezoelectric substance.

Ans. Example of piezoelectric substance. Lead zirconate (pb ZrO3).

2. What type of azerotrope is formed on mixing nitric acid and water?

Ans. On mixing nitric acid and water maximum boiling azerotrope is formed.

3. State the unit of ‘rate constant’ in a zero order reaction.

Ans. Unit of rate constant in zero order reaction is mol L-1 s-1.

4. Why is adsorption always exothermic?

Ans. Due to the bond formation between the adsorbent and the adsorbate.

5. For the reaction A  B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction?

Ans. Order of reaction = ½

6. Why a mixture of Carbon disulphide and acetone shows positive deviation from Raoult’s law? What type of azeotrope is formed by this mixture?

Ans. Intermolecular forces of attraction between carbon disulphide and acetone are weaker than the pure components.Minimumboiling azeotrope at a specific composition.

7. Among the hydrides of Group-15 elements, which have the

i. Lowest boiling point?

ii. Maximum basic character?

iii. Highest bond angle?

iv. Maximum reducing character?

Ans.

i. PH3

ii. NH3

iii. NH3

iv. BiH3

8. A current of 1.50 A was passed through an electrolytic cell containing AgNO3solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow ? (Molar mass of Ag = 108 g mol–1, 1F = 96500 C mol–1).

Ans. Λm = 1000 K/C

Λm =1.65×10−4×10000/0.01 = 16.5 S cm2mol-1

9. How do you convert the following?

i. Ethanal to Propanone

ii. Toluene to Benzoic acid

Ans.

10. 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).

11. Identify the following :

i. Transition metal of 3d series that exhibits the maximum number of oxidation states.

ii. An alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell and lighter flint.

Ans.

i. Mn

ii. Mischmetall

12. Give reasons for the following :

i. Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.

ii. Aquatic animals are more comfortable in cold water than in warm water.

iii. Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

Ans.

i. As compared to other colligative properties, its magnitude is large even for very dilute solutions / macromolecules are generally not stable at higher temperatures and polymers have poor solubility / pressure measurement is around the room temperature and the molarity of the solution is used instead of molality.

ii. Because oxygen is more soluble in cold water or at low temperature.

13. Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol–1, Kf for water = 1.86 K kg mol–1).

Ans. Moles for MgBr2= 10.5/184= 0.0571 mol

Molality = 0.0571200× 1000 = 0.2855 m

i=3

ΔTf= i Kf m= 3× 1.86 × 0.2855 =1.59 K

Freezing point = 273 –1.59 = 271.41 K or -1.59 oC

14. What happens when

i.A freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution ?

ii. Persistent dialysis of a colloidal solution is carried out?

iii. An emulsion is centrifuged?

Ans.

(a) Peptisation occurs / Colloidal solution of Fe(OH)3is formed.

(b) Coagulation occurs.

(c) Demulsification or breaks into constituent liquids.

15. Solve following:

i.Identify the chiral molecule in the following pair :

ii. Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.

iii. Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH.

Ans.

16. 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.

17. Give reason for the following observations :

i. When Silver nitrate solution is added to Potassium iodide solution, a negatively charged colloidal solution is formed.

ii. Finely divided substance is more effective as an adsorbent.

iii. Lyophilic colloids are also called reversible sols.

Ans.

i. The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution.

ii. Due to large surface area

iii. If the dispersion medium is separated from the dispersed phase , the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.

18. Account the following:

i. Write the product when D-glucose reacts with conc.HNO3.

ii. Amino acids show amphoteric behaviour. Why?

iii. Write one difference between -helix and -pleated structures of proteins.

Ans.

i. Saccharic acid / COOH-(CHOH)4-COOH

ii. Due to the presence of carboxyl and amino group in the same molecule / due to formation of zwitter ion or dipolar ion.

iii. α-helix has intramolecular hydrogen bonding while β pleated has intermolecular hydrogen bonding / α-helix results due to regular coiling of polypeptide chains while in β pleated all polypeptide chains are stretched and arranged side by side.

19.A reaction is first order in A and second order in B

i. Write the differential rate equation.

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

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

Ans.

i. Rate = k[A][B]2

ii. Rate becomes 9 times

iii. Rate becomes 8 times

20. Consider the following reaction :

i. Depict the galvanic cell in which the given reaction takes place.

ii. Give the direction of flow of current.

iii. Write the half-cell reactions taking place at cathode and anode.

Ans.

ii. Current will flow from silver to copper electrode in the external circuit.

21. Explain the following:

i. Write the formula of the following coordination compound :

ii. Iron(III)hexacyanoferrate(II)

iii. What type of isomerism is exhibited by the complex [Co (NH3)5Cl] SO4?

iv. Write the hybridisation and number of unpaired electrons in the complex [CoF6]3–. (Atomic No. of Co = 27)

Ans.

i. Saccharic acid / COOH-(CHOH)4-COOH(

ii. Due to the presence of carboxy land amino group in the same molecule / due to formation of zwitter ion or dipolar ion.

iii. α-helix has intramolecular hydrogen bonding while β pleated has intermolecular hydrogen bonding / α-helix results due to regular coiling of polypeptide chains while in β pleated all polypeptide chains are stretched and arranged side by side.

22. Give the formula of monomers involved in the formation of the following polymers :

i.Buna-N

ii. Nylon-6

iii. Dacron

Ans.

23. Mathew works in a multinational company where the working conditions are tough. He started taking sleeping pills without consulting a doctor. When his friend Amit came to know about it he was disturbed and advised Mathew not to do so. He suggested that Mathew should instead practice yoga to be stress free. Mathew is now relaxed and happy after practicing yoga.

After reading the above passage, answer the following questions :

i. Name the class of chemical compounds used in sleeping pills.

ii. Why is it advisable not to take the dose of sleeping pill without consulting a doctor?

iii. Pick out the odd chemical compound on the basis of its different medicinal property: Luminal, Seconal, Phenacetin and Equanil.

iv. List at least two qualities of Amit that helped Mathew to be happy.

Ans.

i. Tranquilizers

ii. It may cause harmful effects and may acts as poison in case of overdose. Therefore, a doctor should be always consulted.

iii. Phenacetin

iv. Empathetic , Caring , sensitive (or any other two relevant values)

24. i. What happens when

(a) Chlorine gas reacts with cold and dilute solution of NaOH ?

(b) XeF2 undergoes hydrolysis?

ii. Assign suitable reasons for the following :

(a) SF6 is inert towards hydrolysis.

(b) H3PO3 is diprotic.

(c) Out of noble gases only Xenon is known to form established chemical compounds.

Ans.

ii. a. Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

b. It contains only two ionisable H-atoms which are present as –OH groups, thus behaves as dibasic acid.

c. Xe has least ionization energy among the noble gases and hence it forms chemical compounds particularly with O2and F2.


OR


i. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F2 and Cl2.

ii. Complete the following reactions :

Ans.

i. a. Fluorine has less negative electron gain enthalpy than chlorine,

b. Fluorine has low enthalpy of dissociation than chlorine

c. Fluorine has very high enthalpy of hydration than chlorine.

d. Fluorine is stronger oxidizing agent than chlorine.

25. i. Give reasons :

a. HCHO is more reactive than CH3-CHO towards addition of HCN.

b. pKa of O2N–CH2–COOH is lower than that of CH3–COOH.

c. Alpha hydrogen of aldehydes & ketones is acidic in nature

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

a. Ethanal and Propanal

b. Pentan-2-one and Pentan-3-one.

Ans.

i. (a) Due to +I effect of methyl group in CH3CHO.

(b) Due to –I effect of nitro group in nitroacetic acid.

(c) Due to the strong electron withdrawing effect of the carbonyl group and resonance stabilisation of the conjugate base.

ii. (a) Add NaOH and I2 to both the compounds and heat, ethanal gives yellow ppt of iodoform.

(b) Add NaOH and I2 to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.

26. (a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).

i. Identify (A) and (B).

ii. Write the structures of (A) and (B).

iii. Why does gas (A) change to solid on cooling?

(b) Arrange the following in the decreasing order of their reducing character:

HF, HCl, HBr, HI

(c) Complete the following reaction:

XeF4+ SbF5 ---------->

Ans.

(a) (i) A = NO2, B = N2O4

(iii) Because NO2 dimerises to N2O4/ NO2 is an odd electron species.

(b) HI> HBr> HCl> HF

(c) XeF4+ SbF5 ----------> [XeF3]+ [SbF6]-


OR


i. Name two carbohydrates which act as biofuels.

ii. What are two major forms of secondary structures in proteins? What kind of bonds stabilize these structures?

iii. What is breeder reactor?

Ans.

i. Two carbohydrates used as biofuels: starch and glycogen.

ii. Regular folding of long peptide chains and coild helix are major forms of secondary structures in proteins.

iii. The reactors where neutrons produced by fission U-235 are partly used to convert less fissionable but abundantly available U-238 and TH-232 into plutonium-239 and uranium-233 are called breeder reactors.

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