07.Sr_Chem

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BOARD OF INTERMEDIATE EDUCATIONSENIOR INTER CHEMISTRY

MODEL PAPER (English Medium)Time : 3 Hours (w.e.f. 2013-14) Max.Marks : 60

SECTION - ANote: (i) Very Short Answer type Questions.

(ii) Answer ALL the question at one place in the same order.

(iii) Each question carries TWO marks. 10 × 2 = 20

1. What are "isotonic solutions"?

2. Λ°m values for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol-1

respectively. Calculate Λ°m for HAc.

3. Give the composition of alloys a) Brass and b) German silver.

4. Aqueous Cu+2 ions are blue in colour, where as aqueous Zn+2 ions are

colourless. Why?

5. Explain the difference between Buna - N and Buna - S?

6. What are "Fibres" and "Elastomers"?

7. What are "antacids"? Give 2 examples.

8. What are "artificial sweetening agents"? Why do we require them?

9. What are "enantiomers"?

10. What are "ambident nucleophiles"?

SECTION - BNote: (i) Short Answer type Questions.

(ii) Answer any SIX questions.

(iii) Each question carries FOUR marks. 6 × 4 = 24

11. Derive Bragg's equation?

12. Define molality. Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in

75 g of benzene (C6H6).

13. Give four differences between characteristics of physical adsorption and

chemical adsorption.

14. Explain the purification of sulphide ore by froth flotation method.

15. How is ammonia manufactured by Haber's process?

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16. Write any four characteristic properties of transition elements.

17. Explain the terms

a) Peptide linkage b) Anomers c) Zwitter ion and d) Vitamins.

18. a) What is "diazotisation"?

b) Give the structures of A, B and C in the following reaction.

CuCN H2O/ H+ NH3C6H5N2Cl → A → B → C

∆

SECTION - C

Note: (i) Long Answer type Questions.

(ii) Answer any TWO questions.

(iii) Each question carries EIGHT marks. 2 × 8 = 16

19. a) State and explain Faraday laws of electrolysis.

b) What is half-life of a reaction?

A first order reaction is found to have a rate constant k = 5.5 × 10-14 s-1. Find

the half life of the reaction.

20. a) Write the names and formulae of the oxo acids of chlorine. Explain these

structures and relative acidic nature.

b) Explain the structures of XeF2 and XeF4.

21. a) What is aspirin? How is it prepared?

b) How ethanol is manufactured from molasses?

c) Explain the terms

i) Cyanohydrin ii) Acetal iii) Semicarbazone and iv) Oxime.



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ANSWERSSECTION - A

1. What are "isotonic solutions"?

Ans: Two different solutions having same osmotic pressure at a given temperature

are called "isotonic solutions".

We.g.: Blood is isotonic with 0.9% NaCl solution.

V

2. Λ°m values for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2

respectively. Calculate Λ°m for HAc.

Ans: Λ°m

(HAc) = Λ°m (HCl) + Λ°

m (NaAc) - Λ°m (NaCl)

= 425.9 + 91.0 - 126.4 = 390.5 S cm2 mol-1.

3. Give the composition of alloys a) Brass and b) German silver.

Ans: a) Brass 60% Cu & 40% Zn.

b) German silver: 25 - 40% Cu, 40 - 50% Ni, 25 - 35% Zn.

4. Aqueous Cu+2 ions are blue in colour, where as aqueous Zn+2 ions arecolourless. Why?

Ans: Cu+2 = [Ar]4s0 3d9 Zn+2 = [Ar]4s0 3d10

Due to presence of one unpaired d electron in 3d sub shell of Cu+2, due to

d - d transition, it absorbs yellow colour and transmits blue colour. So aqueous

Cu+2 ions are blue in colour. As Zn+2 does not contain unpaired d electrons,

aqueous Zn+2 ions are colourless.

5. Explain the difference between Buna - N and Buna - S?

Ans: Buna - N: Copolymer of 1, 3 Butadiene & Acrylonitrile.

Buna - S: Copolymer of 1, 3 Butadiene and Styrene.

6. What are 'Elastomers" and "Fibres"?

Ans: Elastomers: Polymers in which the inter molecular forces of attraction

between the polymer chains are weakest.

e.g.: Buna - S, Buna - N.

Fibres: Polymers in which the intermolecular forces of attraction between the

polymer chains are strongest due to Hydrogen bonding (or dipole -dipole interaction).

e.g.: Polyester, Nylon 6, 6.



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7. What are "Antacids"? Give 2 examples.

Ans: The chemical substances that removes excess of acid by neutralisation by

bringing pH to the normal level in stomach is called "Antacid".

e.g.: Omeprazole, Lansoprazole.

8. What are "artificial sweetening agents? Why do we require them?

Ans: The chemical substances not only controls the intake of calories but also gives

sweet taste are called artificial sweetening agents.

e.g.: Sucralose, Saccharin.

9. What are "Enantiomers"?

Ans: A pair of stereoisomers having non - super impossible mirror images are called

"Enantiomers".

e.g.: (+) lactic acid & (-) lactic acid.

10. What are "ambident nucleophiles"?

Ans: Species having 2 nucleophilic centres are called ambident nucleophiles".

− −. . . . . .e.g.: Cyanide : C ≡ N:, Nitrate :O − N = O :. .

SECTION - B11. Derive Bragg's equation.

Ans: A crystal has many planes.

Atoms or ions are arranged in systematic

geometry in these planes. When X - rays

incident on these crystal planes, they undergo

diffraction. When the waves are diffracted

from these atoms or ions, they may have

constructive interference or destructive

interference. From the figure, it is very clear that 1st and 2nd X-rays

travel the same distance till the wave front AD. Whereas 2nd X-ray travels an

extra distance DB + BC (path difference) than that of 1st X-ray. If the two

waves are present in the same phase (constructive interference), the path

difference must be equal to the wavelength λ or an intergral multiple of wave

length.



θD C

B

A

1st ray

2nd ray

θ↑d↓

1st plane

2nd plane

3rd plane

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DB DBIn ADB traingle, sin θ = = ∴ dB = d sin θ

AB d

BC BCIn ABC traingle, sin θ = = ∴ BC = d sin θ

AB d

∴ nλ = DB + BC = 2d sin θ.

This relation is known as Bragg's equation.

θ values increases with n values. By knowing the values of θ and λ one can

calculate d.

λ = wavelength of X-ray

θ = angle of incident X-ray

n = order of diffraction

d = interplanar distance.

12. Define molality. Calculate molality of 2.5 g of ethanoic acid (CH3COOH)

in 75 g. of benzene (C6H6).

Ans: Number of gram moles of solute present in 1 kg of solvent is called molality.

weight of solute 1000Molality (m) = ×

gram molar weight of solute weight of solvent (grams)

2.5 1000= ×

60 75

= 0.55 moles/ kg.

13. Give 4 differences between characteristics of physical adsorption and

chemical adsorption.

Ans: S.No. Physical Adsorption Chemical Adsorption

1. Enthalpy of adsorption is low. Enthalpy of adsorption is high.

2. It is multilayered. It is unilayered.

3. It is not specific. It is highly specific.

4. It is reversible. It is irreversible.



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14. Explain the purification of sulphide ore by froth flotation method.

Ans:

★ This process is used to concentrate sulphide ores.

★ Finely powdered ore is taken into a tank along with water.

★ Froth collectors like pine oil, sodium ethylxanthate are added for

non-wettability of ore particles.

★ Stabilizers like cresols, aniline are added to stabilize the froth.

★ This suspension is agitated with a rotating paddle and air in order to form

froth.

★ The ore particles are wetted by oils and the gangue particles are wetted by

water.

★ The froth is separated and dried to recover the ore particles.

15. How is ammonia manufactured by Haber's process?

Ans:



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According to Lechatlier's principle NH3 is manufactured as follows:★ In this method pure, dry N2 & H2 are taken in 1 : 3 ratio and compressed at

200 atm with compressed pump.

★ After drying, the mixture is passed through catalytic chamber, where Al2O3& K2O mixture is taken as catalyst.

★ 700 to 773 K temperature is maintained to get 10% NH3.

700 - 773 KN2 (g) + 3 H2 (g) 2 NH3 (g)

200 atmAl2O3 + K2O

★ NH3 is dried by using quick lime (CaO).16. Write any four characteristic properties of transition elements.

Ans: ★ They form alloys readily, due to similar atomic radii (differ not more than15%).

★ They form coloured compounds due to presence of incomplete d-subshelland d - d transition.

★ They exhibit variable oxidation states as both (n - 1)d and ns electrons takepart in bonding (due to little difference in the energies of these orbitals).

★ They form complex compounds due to smaller atomic size, high ioniccharge and availability of vacant d-orbitals.

17. Explain the terms a) Peptide linkage b) Anomer c) Zwitter ion andd) Vitamins.

Ans: a) Peptide linkage: A peptide linkage (or bond) is formed when -NH2 of oneamino acid reacts with -COOH of another amino acid.

-CO OH + -H NH → -CONH- + H2OPeptide linkage

b) Anomers: Anomers are the stereoisomers which differ in their configurationat C - 1.

e.g.: α - D - glucopyranose & β - D - glucopyranose.

c) Zwitter ion: A dipolar neutral ion formed when the carboxyl group loses aproton and this proton is gained by amino group in aqueous solution of amino acid iscalled Zwitter ion.

e.g.: R - CH - COO-N+H3

d) Vitamins: Naturally occuring organic compounds which are required insmall quantities for the maintenance of normal health and growth of organism.

e.g.: Vitamin A, Vitamin C.



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18. a) What is" diazotisation"?

b) Give the structures of A, B and C in following reaction

CuCN H2O/ H+ NH3C6H5N2Cl →→ A →→ B →→ C

KCN ∆∆ ∆∆Ans: a) Diazotisation: The conversion of aromatic primary amines into diazonium

salts at 0 - 5°C, by the reaction of aromatic amine with nitrous acid (NaNO2+ HCl) is called "diazotisation".

NH2 N2Cl

0 - 5°C + NaNO2 + 2 HCl → + NaCl + 2 H2O

CuCNb) C6H5N2Cl → C6H5CN + N2

KCN (A)H2O/ H+

C6H5CN → C6H5COOH(B)

∆C) C6H5COOH + NH3 → C6H5CONH2+ H2O

(C)

SECTION - C19. a) State and explain Faraday laws of electrolysis.

b) What is half life of a reaction?

A first order reaction is found to have a rate constant k = 5.5 × 10-14 s-1.Find the half life of the reaction.

Ans: Faraday's first law: The mass of a substance (m) deposited (or dissolved or

liberated) at an electrode during electrolysis is directly proportional to the

quantity of electricity (Q) passed through fused or aqueous electrolyte.

m ∝ Q Where e = electrochemical equivalent.

m ∝ I.t I = current in amperes

m = e.I.t t = time in seconds.

Faraday's second law: When the same quantity of electricity is passed through

different cells containing fused or aqueous electrolytes, connected in series, the

mass of substances deposited (or dissolved or liberated) at electrodes or in the

ratio of their equivalent weights.

e.g.: mCu : mAg : mH2= ECu : EAg : EH2



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(b) Half life period: The time in which the concentration of a reactant is

reduced to half of its initial concentration.

0.693 0.693t 12

= = = 1.26 × 1013 sec.k 5.5 × 10-14

20. (a) Write the names and formulae of the oxo acids of chlorine. Explaintheir structures and relative acidic nature.

(b) Explain the structures of XeF2 and XeF4.

Ans: (a) Oxo acids of Chlorine:

S. Formula Name of the Type of shape

No. oxo acid hybridisation

. .1. HClO Hypochlorous acid sp3 :Cl − OH. .

Linear

: :2. HClO2 Chlorous acid sp3 Cl Angular

O OH

. .3. HClO3 Chloric acid sp3 Cl

O O OH

Pyramidal

O

4. HClO4 Perchloric acid sp3 Cl

O O OH

Tetrahedral

Relative acidic nature: HOCl < HClO2 < HClO3 < HClO4.

(b) XeF2:

Xe = 5s2 5px2 5py

2 5pz1 5d1

(First excited state)

Xe undergoes sp3d hybridisation.

The shape of XeF2 is linear.



F

F

:

:

:

Xe

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XeF4:

Xe = 5s2 5px2 5py

1 5pz1 5d1 5d1

(Second excited state)

Xe undergoes sp3d2 hybridisation. Due to presence of 2 lone pairs, 4 bond pairs

in Xe. The shape of XeF4 is square planar.

21. (a) What is Aspirin? How is it prepared?

(b) How ethanol is manufactured from molasses?

(c) Explain the terms

(i) Cyano Hydrine (ii) Acetal (iii) Semicarbazone and (iv) Oxime.

Ans: (a) Aspirin: Acetyl salicylic acid is called Aspirin. Salicylic acid on acetylation

with acetic anhydride gives Aspirin.

COOH COOH

H+ COOCH3O H + CH3COO COCH3 → + CH3COOH

(b) Ethanol from molasses: Diluted molasses (10% sucrose) is taken with

H2SO4 (pH = 4) and Yeast in fermentation tank.

Invertase C12H22O11 + H2O → C6H12O6 + C6H12O6

Sucrose Glucose Fructose

Fermentation takes place in anaerobic conditions (in absence of air) and CO2 is

released and C2H5OH is formed in presence of Zymase enzyme.

Zymase C6H12O6 → 2 C2H5OH + 2 CO2

Glucose/ Fructose

"Wash produced in this step on distillation gives finally absolute (100%)

alcohol.



::

Xe

F

F F

F

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(c) (i) Cyanohydrin: Acetaldehyde reacts with HCN to give acetaldehydecyanohydrine.

O OH

H3C C + HCN → H3C C CN H H

(ii) Acetal: Acetaldehyde on reaction with 2 moles of CH3OH in presence of

dry HCl gives acetal.

H3C Dry HCl H3C OCH3C = O + 2 CH3OH → C

H H OCH3

(iii) Semicarbazone: Acetaldehyde on reaction with semicarbazide givesacetaldehyde semicarbazone.

H3C C = O + H2 N NHCONH2 → H3C C = N - NHCONH2 H H

(iv) Oxime: Acetaldehyde reacts with NH2OH to give acetaldoxime.

H3C C = O + H2 N - OH → H3C C = N OH + H2O H H

- A.N.S.Sankara Rao, Senior Lecturer.



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