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COUNCIL OF HIGHER SECONDARY EDUCATION, MANIPUR |
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SAMPLE QUESTION, 1999 |
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CHEMISTRY |
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( THEORY ) |
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Full Mark - 70 |
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Pass Mark - 21 |
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Time : 3 hours |
(i) There are 36 questions in all.
(ii) All questions are compulsory.
(iii) Questions of numbers 1 to 12 are of very short answer type of 1 mark each.
(iv) Questions of numbers 13 to 17 are of objective type. Each question carries 1 mark and out of the four choices given, only one is correct. Rewrite the question number and the correct answer in the answer script.
(V) Questions of numbers 18 to 27 are of short answer type of 2 marks each. each.
(vi) Questions of numbers 28 to 33 are of short answer type of 3 marks each.
(vii) Questions of numbers 34 to 36 are of long answer type of 5 marks
(viii) Marks are indicated at the right side margin of each question.
(ix) Use of long table is allowed.
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Council of Higher SecondaryEducation, Manipur
HIGHER SECONDARY EXAMINATION 1999
CHEMISTRY : (THEORY)
Full Marks-70
MARKII\IG SCHEME/SCORING KEY/INSTRUCTION TO EXAMINERS
1. Statement of Hess’s law of constant heat of summation. 1
2. B describes the shorter wave length. 1
=h/p
3. Unit of equivalent conductance given is wrongand therefore, full marks will be awarded to all. 1
4. Definition of activation energy. 1
5. Zero order. [rate is independent of concentration 1
6. CH3C COOH . H
7. Any double salt e.g. FeSO4 (NH4 ) 2 SO4.6H2O. 1
8. Finely divided Powdered catalyst has more surface area and hence more active sites.
9. -CH2-CH-Cl
10. The C-O-C linkage between two components of an acetal is the glycoside linkage. 1
11. Pheromones are the insect/some animal hormones which are principal means of communication among themselves. 1
12. Malathion is biodegradable while DDT and Aldrin are eot biodegradable. Hence malathion is more safely used.
13. (a) O2- 1
14. (d) F 1
15. (c) HCOOCH3
16. (a) 500 years. 1
17. (b) dE+w=0 1
18. B.C.C. with Na+ at the centre of the cube.Drawing...........................................
Labelling.......................................
No mark should be awarded for drawing if laballing is wrong.
19. AGO = - 2.303 RT log K.=-2.303x8.3 (JK-l mol-1) x 298 (K) x log 2 x 10-7
=38.159 (KJ mol-1) Correct formula 1
Correct calculation 1
20. (a) It is a process of photochemical reaction inwhich one of the reactants absorb a photon from the light forming reactive species. 1
b) photosensitization is the process of activation of a reactant which itself is not photoactive, but do so by absorbing lightenergy via a foreign photoactive substance and proceeds the reaction. 1 2
21. C6H5OH or -OH Ionised as- CH -O(-)+H(+) Whereas C2H5OH does not. It is because of fact that C2H5O(-) is not stabilised by resonance while 0(-) ion is stabilisen by resonance as follows : 2
22. Lanthanide contraction, statement. 1Any two factors at minimum. 1/2x2
23. (i) Pentaamminebromoiron (11) sulphate 1/2
(ii) Pentaamminesulphatoiron (11) bromide 1/2
The former will be precipitated with BaC1 2solution. 1 2
24. Outline diagram of any design of the apparatusof electro-osmosis 1 Labelling 1 2
25. Definition of Polyamide. 1
Synthesis of nylon - 66 (may be in words or reaction or both) 1
26.C)Adenosine (nucleoside) 11/2Phosphate group is attached to C5 or C3 of thedeoxyribose part of the nucleoside. 1/2
27. The oxygen of air inhaled in respiration is used as oxidant in the biological processes. 1Oxygen taken by the body through lungs is carried to the tissues by blood. 1 2
28.Definition of atomic orbital. 1
The azimuthal quantum number is related to the angular momentum of the electron. Thus it icads to the different shapes of the orbitals. 1Statement of Pauli’s exclusion principle. 1
29. (i) Pt | Fe+2 (aq).Fe+3 (aq)| Br-(aq) | ½ Br2 (aq)| pt. 1
(ii) Fe +2 (aq)+1/2Br2(aq)= Fe+3(aq)+Br-(aq) 1
(iii) E cell=EO cell-2:@@
11log [F (aa). [Br-(aq)j [Fe (.q)][Br2)aq)]vlWhere E0cell=(]. 1 7) & n - 1.
30. Haloalkanes are the halogen derivatives of alkanes. 1
Any method of preparation 1
Any synthetic use 1 3
31. (a) CH3>C=O+H2N-OH--.>. CH,, > C -N-OH+H,10CHS CH3
(b) CH " CO <O+NI-I3-->CH3CONH,+CHaCOONH4 CH3C0
(C) C6 H 5 CO OH + SOCI2 C6H5 COCI +SO2 + HCI. 1+1+1 3
32. (a) K2cr207 +4Nacl+6 conc. H2SO42KHSO4+2cr O2Cl2+4NaHSO4+ 3H2O. NaC1 may be replaced by any other chloride. Orange red vapours ofCro2C12 shows the presence of chloride ion. 1
(b) 2KMnO4+8H2SO4+10 FeSO4 K5SO4+2 MnSO4+5Fe2(SO4)3+8H2O 1
(c) Due to the presence of completly filled inner d- orbital. 3
33. P-ray207
(i) M is pb 182
136(ii) N is Sn
50 1 3
34. Elements having electronic configuration ns2np1-6 are P-block elements. Or, any other suitable statement may be given. 1
Elements of group 17: F, C1, Br, I, At. 1
At. is naturally radioactive 1
Suitable explanation. 2 5
35. A- (benzene) 1
B- NO2 (nitrobenzene) 1
C- NH2 (aniline) 1
D- N=N+CI(benzene diazonium chloride) 1
E- CI (chlorobenzene) 1 5
35. Lowering of V.P. : Suitable explanation. 1
Raoult’s law : L p /PO=X2
where X2 is the mol-fraction of the solute. 1
Deduction of W2yMI
M2 = EP for very dilute solution.WI Ml=molar mass of solvent M2=molar mass of soluteW2=mass of solvent in gramW2=mass of solute in gram. 3 5
Sd/ Head Examiner