CSSC CHEMISTRY MARKING SCHEME
SET -3
1. 1
C) (As the value of standard reduction potential decreases
the reducing power increases)
2. c) 1
concentration vs time
3. 1
d) i-B ii-A iii-D iv_C
4. 1
a) drops of HCl
5. 1
b) 40
6. d)2,6 1
7.
c) no unpaired electron, zero spin 1
8. d ) Haemoglobin
1
9. c) Ethanol < phenol< acetic acid < chloroacetic acid 1
10 a) Allyl 1
11 (a) (i) and (ii)
1
12 c) A- Pyridinium chlorochromate B- Acidified KMnO4 1
13
(c) Assertion is true but Reason is false.
1
14 c) Assertion is true but Reason is false
1
15 (c) Acetylation decreases the electron-density in the benzene ring thereby
preventing oxidation. 1
16 a) R is the correct explanation of A. Due to +M effect of 1
OH .its intermediate carbocation is more stable than
the one in benzene
17 a) amylose and amylopectin ½
b) Starch – alpha C1 C4 linkage. ½
Cellulose – C1 C4 beta linkage ½
c) polysaccharides of glucose . form of energy storage in fungi and ½
animals.
d)storage form of glucose in animals. Found in liver.
a)Equation Toluene 1+1
18. b) Equation 1- ethylcyclohex – 1- ene
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19. 1
ΔTb = i kb m
i= 2
m = 4/120 x 10 = 1/3
= 2x 0.52 x 1/3 = 0.347
Boiling point of solution = 100 + 0.347 = 100.347 C
OR
Formula
0.850 – 0.845/ 0.850 = 0.5 x 78/39 x M2
M2 = 170 g/mol
20. 1
ΔG = -2 x 96500 x 0.236
-45 .55 kJ /mol
Lok Kc = -ΔG0 / 2.303 RT 1
= -45.55 / 2.303 x 8.314 x 10-3 x 298
= 7.983.
21. 1
a) [Co (NH3)4Cl2] Cl
b) optical isomerism 1
22. 1
a)A = Butanal
B = Butan – 2 -one 1
C = 2- methylpropanal
D = Butane
b) B 1
Or
a) HCHO – Fehlings reddish orange ppt 1
Benzaldehyde – will not answer fehlings
b) Any suitable method
1+1
23. 1
a) 2 – chloro 2 methylpropane SN1
b) b)1-Bromo-2-methylbut-2-ene 1
c) 1
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24. a) increases conductivity of electrolyte and ammonia gas 1
b) 425.9 + 91 – 126.4 = 390.5 S cm2mol-1 1
c) Equations 1
25 a) E0 cell = 0.34 – (-1.66) = 2 v 1
n=6
E cell = 2- 0.059 /6 log[0.15]2 /( 0.025)3
1
2- 0.059/6 (2 log 0.15 – 3 log 0.025)
1
1.9689 V
b) H2 at Cathode and Cl2 at anode
26 a) metal atom has some electrons which are not involved in bonding 1
and hence it can donate electrons .
Higher oxidation state – no electrons available for donation , so it can 1
accept electron.
1
b) Cr – Presence of unpaired electron – strong metallic bonding
Hg – no unpaired electron – weak metallic bonding
c)shielding effect of n-1 d electrons (steady increase in effective
nuclear charge is counterbalance)
27. a) aldehyde + reaction ½+½
b) Starch 1
c) phosphodiester linkage 1
28. a) LiAlH4 1
b) ortho nitrophenol – intramolecular H bonding -more volatile 1
para nitrophenol – intermolecular H bonding – less volatile
1
c) Benzoquinone (write equation)
29. 1
a) (i) (C2H5)3N, (C2H5)2NH, C2H5NH2, NH3
(ii) (C2H5)2NH, (C2H5)3N, C2H5NH2, NH3 1
b) Aniline is less basic than ammonia because the lone pair of electrons on
the nitrogen atom in aniline is delocalized over the benzene ring, making it
less available for donation / protonation
c)
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1
, anilinium chloride
(OR)
c) A: [CH3NH3]+ OH-- B: FeCl3
1
30. 1
a (a)
(i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium
1
chloride solution or blood cells thus, on placing blood cells in this solution
exosmosis takes place that results in shrinking of cells.
(ii) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium
chloride solution or blood cells thus, on placing blood cells in this solution
endosmosis takes place that results in swelling of cells.
(b) macromolecules as they are generally not stable at higher temperatures
and polymers have poor solubility
(c) W2 10.50 g, W1 200 g
1+1
M 2 MgBr 2 = 184 g mol -1
K f 1.86 K kg mol-1
MgBr Mg 2 2 Br , i = 3
2 aq aq
aq
i K f W2 1000
TF iK f m, T f
M 2 W1
3 1.86 10.50 1000
T f 1.592 K
184 200
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Freezing point of solution, T f T f T f = 273 – 1.592 = 271.408 K
(OR)
Give, weight of urea W2 = 30 g 1 +1
Weight of water W1 = 846 g
Vapour pressure of water p1 = 23.8 mm Hg
30 846
nB 0.5, n A 47
60 18
nA
Mole fraction of water x A
47 47
= 0.99
n A n B 47 0.5 47.5
31. ½ x4
a)(i) =2
1
(ii)
2
BROMINE WATER reagent observation
TEST
Phenol Bromine water White precipitate due 1+ 1+
to the formation of 1
2,4,6 tribromophenol 1
1
Benzyl alcohol Bromine water NO REACTION
(OR ANY OTHER SUITABLE TEST)
(III) + C2H5OH
(tert.butyl iodide) (ethanol)
(b)(i) Only one of the two -NH2 groups in semicarbazide is involved in the
formation of semicarbazones because the lone pair of electrons over the
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other NH2 group is involved in conjugation with >C=O group and is not
available for nucleophilic attack
(ii) The carboxylic carbon is less electrophilic than carbonyl carbon
because of the possible resonance
structure shown below:
OR
a) Equations
b)Phenol – neutral FeCl3 – violet colour
Ethanol – no reaction with neutral FeCl3 (or any other test)
c) Lower iodide and higher alcohol – CH3Br + CH3 CH2CH2 OH
32.
OR
(a) (i) due to small size, high change to size ratio & availability of vacant
‘d’orbitals.
1
1
(ii)Highest oxide is covalent & lowest oxide is ionic.
1
(iii) Mn2+ (has d5 – stable) but for chromium Cr3+ (has t2g3 – stable)
1
(b) Electronic configuration + Oxidation state
Or
1
a) i) [CoCl H2O (NH3) 4] Cl2
1
ii) [ Co(en)3] Chelate ligand
1
iii) EDTA DMG
½x4
=2
b) Cr+3
Mn+3
Ti+4
Mn+3
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33. 1
a) P1 0 - p1/ p1 0 = WB x Ma/ Mb x WA 1
23.8 – p1/23.8 = 30 x 18/60x 846 = 23.55 mm Hg 1
b) 2 Differences 1+1
OR
a) 273.15 – 269.15 = kf x10 x1000/ 342 x 90 1
kf = 12.3 1
ΔTf = kfx m
= 7.6 1
Tf = 273.15 – 7.6 = 265.55 K 1
1
b) Definition
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