CSSC CHEMISTRY MARKING SCHEME- SET -1

1. 1
C(As the value of standard reduction potential decreases
the reducing power increases)
2. c) (i), (iii) and (iv) 1

3. 1
A HCHO
4. 1
C Concentration Vs Time
5. 1
d i-B ii-A iii-D iv_C
6. A drops of hydrochloric acid 1

7. B 40 min
1
8. D [2,6]
1
9. C no unpaired electron, zero spin 1

10 D Haemoglobin 1

11 c)Ethanol < phenol< acetic acid < chloroacetic acid

1
12 A allyl 1

13 b
1
14 a
1
15 d
1
16 b 1

17 a) amylose and amylopectin ½

b) Starch – C1 C4 linkage. (α D glucose) ½
Cellulose – C1 C4 linkage (βD glucose) 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 product 1+1
18. b) Equation 1- ethylcyclohex – 1- ene

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

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 -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
i) Any suitable method 1
ii) Ethyl Benzene + alk KMnO4 → Benzoic acid 1

23. 1
a) 2 – chloro 2 methylpropane SN1
b) (i) Chiral, non-superimposable lacks plane of symmetry 1+1

24. a) increases conductivity of electrolyte and ammonia gas 1

b) 425.9 + 91 – 126.4 = 390.5 S cm2mol-1 1
c) A: Pb + SO42_ → PbSO4 + 2e- 1
C: PbO2 + SO42_ + 4H+ + 2 e- → PbSO4 + 2 H2O

25 K2 Ea T2  T1  1
log =
K 1 2.303 R  T1T2 

24  10 2 Ea  50 
log = 1
4  10 2
2.303  8.314  350  300 
Ea
log 6 =
19.147  7  300
Ea = 0.7782  19.147  7  300
= 31290.41 J/mol.
= 31.29 KJ / mol 1

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

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 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. a) electron donating group , ortho and para electron rich .Draw 1
resonating structures
1
b) intermolecular H bonding

c) Anilinium ion is meta directing . 1+ 1

1 > 2> 3

OR
NH2 group is acetylated so that controlled nitration can occur at the
para position + Reason 1+1
Flurobenzene

30. 1
a) Write expression rate if disappearance of N2 ( show decrease in
conc N2 per time)
1
b) K is rate of reaction when concentration of reactant is unity

c) ¼ times l mol=1 s-1 1+1

OR
Definitions
1+ 1

31.
a) i)CH3CH(OH)CH2CH3

ii) Ortho and para bromo acetophenone structures

1

b) (A) (CH3)2CHCH2Br 1

(B) (CH3)2CH=CH2

(C) (CH3)3CBr and equations

OR
a) Equations
b)Phenol – neutral FeCl3 – violet colour

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 Ethanol – no reaction with neutral FeCl3 (or any other test)

c) Lower alkyl iodide and higher alky;l alcohol – CH3Br + CH3

CH2CH2 OH

1
32. a) (a) In aqueous solution, CuCl2 ionises into
CuCl2 Cu2+ + 2Cl–
At cathode : Cu2++ 2e-Cu 1
At anode : 2Cl– -Cl2 + 2e–
At the cathode, the copper ion will be deposited because it has a higher 1
reduction potential than the water molecule.
At the anode, the lower electrode potential value will be preferred but due
to over potential of oxygen, chloride ion gets oxidized at the anode.
(b) Given cell,

2 Al s   3Cu 2 0.01M   2 Al 3 0.01M   3Cu  s 

1

E  = 1.98 V, E  = ?
cell cell

Using Nernst equation at 298 K

2 1
 Al 3  
 
 E  log 
0.0591
E
cell cell 6 3
Cu 2  
 

1.98 V = E  
0.0591
log
10  2  2

cell 6 10  2  3

1.98 V = E 
0.0591
  log 102
cell 6

1.98 V = E 
0.0591
 2
cell 6

E  = 1.98 
0.0591 1
 1.99 V
cell 6

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 OR
a) i) octahedral Paramagnetic
ii) Cis
iii) t2g4 , eg2

b) i)Mn+2 = 3d5 high ioniosation enthalpy and low hydration enthalpy

for Cu+2/Cu

ii)MnO4-

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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