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CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
CBSE
Class XII Chemistry
Sample Paper 2 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. Due to intermolecular hydrogen bonding in alcohols, which is absent in ethers, alcohols
have a higher boiling point than ethers.
2. Polymerisation: It is the process in which a number of monomers join together to form
a long-chain polymer.
2
O ,350 570K
2 2 2 2 n 1000 2000atm
Ethene Polyethene
(monomer) (polymer)
nCH CH (CH CH )
?
?
? ? ? ? ? ? ? ? ? ?
3.
1
8(corneratom)× atomperunit cell=1
8
OR
When NaCl is doped with MgCl2, each Mg
2+
ion replaces 2 Na
+
ions to maintain electrical
neutrality. One site is occupied by Mg
2+
ion, while the other remains vacant. Thus, cation
vacancies are formed. This defect is called an impurity defect.
4. Milk is an example of an oil in water emulsion, and cod liver oil is an example of a water
in oil emulsion.
5. The given metal is a divalent ion = M
2+
. It has atomic number 25 and will have d
5
electronic configuration.
?The magnetic moment µ is
µ = 5 ( 5 + 2 ) = 5 . 9 2B M
OR
The order of increasing oxidising power is
7
2
2 2 4
VO Cr O MnO
? ? ?
??
Section B
6.
B
T 80.31 C 80.10 C 0.21 C ? ? ? ?
WB= 1.25 g, Kb =2.53 ?C kg mol
1-
, MB =?, WA = 99 g
bB
B
bA
B
K ×W
M = ×1000
?T ×W
2.53×1.25×1000
=
0.21×99
M =152g/mol
Page 2
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
CBSE
Class XII Chemistry
Sample Paper 2 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. Due to intermolecular hydrogen bonding in alcohols, which is absent in ethers, alcohols
have a higher boiling point than ethers.
2. Polymerisation: It is the process in which a number of monomers join together to form
a long-chain polymer.
2
O ,350 570K
2 2 2 2 n 1000 2000atm
Ethene Polyethene
(monomer) (polymer)
nCH CH (CH CH )
?
?
? ? ? ? ? ? ? ? ? ?
3.
1
8(corneratom)× atomperunit cell=1
8
OR
When NaCl is doped with MgCl2, each Mg
2+
ion replaces 2 Na
+
ions to maintain electrical
neutrality. One site is occupied by Mg
2+
ion, while the other remains vacant. Thus, cation
vacancies are formed. This defect is called an impurity defect.
4. Milk is an example of an oil in water emulsion, and cod liver oil is an example of a water
in oil emulsion.
5. The given metal is a divalent ion = M
2+
. It has atomic number 25 and will have d
5
electronic configuration.
?The magnetic moment µ is
µ = 5 ( 5 + 2 ) = 5 . 9 2B M
OR
The order of increasing oxidising power is
7
2
2 2 4
VO Cr O MnO
? ? ?
??
Section B
6.
B
T 80.31 C 80.10 C 0.21 C ? ? ? ?
WB= 1.25 g, Kb =2.53 ?C kg mol
1-
, MB =?, WA = 99 g
bB
B
bA
B
K ×W
M = ×1000
?T ×W
2.53×1.25×1000
=
0.21×99
M =152g/mol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
OR
Given:
i = 3,
f
K = 1.86 K kg mol
-1
,
B
W = 10.5 g,
B
M = 184 g mol
-1
,
A
W = 200 g
fB
f
BA
i×K ×W ×1000
? T =
M ×W
On substituting the values in the above equation, we get
f
f
o
f f f
f
3×1.86×10.5×1000
? T = = 1 . 5 9 K
184×200
Aswe know ? T , l e t s f ind f r e e z ing point o f a q u e o ussolution,
T =T - ?T
T =273.15K-1.59K=271.56K
7. The resonating structures of NO2 are
The resonating structures of N2O2 are
8.
3 2 2 7 2 4
CaCO K Cr O /H SO
3 2 3 3 2 3 3 Oxidation
Ethanol Ethanoicacid Ca.Ethanoate Ethanol
CH CH OH CH COOH (CH COO) Ca CH COCH ? ? ? ? ? ? ? ? ? ? ? ? ?
9.
(aq )
(s) 2(s)
Cd 2OH Cd(OH) 2e
??
? ? ? ? ?
2(s) 2 2(s)
NiO +2H O 2e Ni(OH) +2OH
??
? ? ? ?
10.
(a) Molecularity: Pseudomolecular
(b) Order of reaction: First order (1)
11.
(a) When phenol is treated with FeCl3 solution, it gives violet colouration. Cyclohexanol
does not give any colouration.
3
6 5 3 6 5 6
Phenol Violet colouration
6C H OH FeCl [Fe(OC H ) ] 3H 3HCl
??
? ? ? ? ? ?
Cyclohexanol
Page 3
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
CBSE
Class XII Chemistry
Sample Paper 2 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. Due to intermolecular hydrogen bonding in alcohols, which is absent in ethers, alcohols
have a higher boiling point than ethers.
2. Polymerisation: It is the process in which a number of monomers join together to form
a long-chain polymer.
2
O ,350 570K
2 2 2 2 n 1000 2000atm
Ethene Polyethene
(monomer) (polymer)
nCH CH (CH CH )
?
?
? ? ? ? ? ? ? ? ? ?
3.
1
8(corneratom)× atomperunit cell=1
8
OR
When NaCl is doped with MgCl2, each Mg
2+
ion replaces 2 Na
+
ions to maintain electrical
neutrality. One site is occupied by Mg
2+
ion, while the other remains vacant. Thus, cation
vacancies are formed. This defect is called an impurity defect.
4. Milk is an example of an oil in water emulsion, and cod liver oil is an example of a water
in oil emulsion.
5. The given metal is a divalent ion = M
2+
. It has atomic number 25 and will have d
5
electronic configuration.
?The magnetic moment µ is
µ = 5 ( 5 + 2 ) = 5 . 9 2B M
OR
The order of increasing oxidising power is
7
2
2 2 4
VO Cr O MnO
? ? ?
??
Section B
6.
B
T 80.31 C 80.10 C 0.21 C ? ? ? ?
WB= 1.25 g, Kb =2.53 ?C kg mol
1-
, MB =?, WA = 99 g
bB
B
bA
B
K ×W
M = ×1000
?T ×W
2.53×1.25×1000
=
0.21×99
M =152g/mol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
OR
Given:
i = 3,
f
K = 1.86 K kg mol
-1
,
B
W = 10.5 g,
B
M = 184 g mol
-1
,
A
W = 200 g
fB
f
BA
i×K ×W ×1000
? T =
M ×W
On substituting the values in the above equation, we get
f
f
o
f f f
f
3×1.86×10.5×1000
? T = = 1 . 5 9 K
184×200
Aswe know ? T , l e t s f ind f r e e z ing point o f a q u e o ussolution,
T =T - ?T
T =273.15K-1.59K=271.56K
7. The resonating structures of NO2 are
The resonating structures of N2O2 are
8.
3 2 2 7 2 4
CaCO K Cr O /H SO
3 2 3 3 2 3 3 Oxidation
Ethanol Ethanoicacid Ca.Ethanoate Ethanol
CH CH OH CH COOH (CH COO) Ca CH COCH ? ? ? ? ? ? ? ? ? ? ? ? ?
9.
(aq )
(s) 2(s)
Cd 2OH Cd(OH) 2e
??
? ? ? ? ?
2(s) 2 2(s)
NiO +2H O 2e Ni(OH) +2OH
??
? ? ? ?
10.
(a) Molecularity: Pseudomolecular
(b) Order of reaction: First order (1)
11.
(a) When phenol is treated with FeCl3 solution, it gives violet colouration. Cyclohexanol
does not give any colouration.
3
6 5 3 6 5 6
Phenol Violet colouration
6C H OH FeCl [Fe(OC H ) ] 3H 3HCl
??
? ? ? ? ? ?
Cyclohexanol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
(b) In the iodoform test, isopropyl alcohol is warmed with NaOI (I2NaOH) to give a
yellow precipitate of idoform, while benzyl alcohol does not give any yellow
precipitate.
12. Addition polymer: polyvinyl chloride, polyethene
Condensation polymers: terylene, Bakelite
Section C
13.
(a)
(b)
Page 4
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
CBSE
Class XII Chemistry
Sample Paper 2 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. Due to intermolecular hydrogen bonding in alcohols, which is absent in ethers, alcohols
have a higher boiling point than ethers.
2. Polymerisation: It is the process in which a number of monomers join together to form
a long-chain polymer.
2
O ,350 570K
2 2 2 2 n 1000 2000atm
Ethene Polyethene
(monomer) (polymer)
nCH CH (CH CH )
?
?
? ? ? ? ? ? ? ? ? ?
3.
1
8(corneratom)× atomperunit cell=1
8
OR
When NaCl is doped with MgCl2, each Mg
2+
ion replaces 2 Na
+
ions to maintain electrical
neutrality. One site is occupied by Mg
2+
ion, while the other remains vacant. Thus, cation
vacancies are formed. This defect is called an impurity defect.
4. Milk is an example of an oil in water emulsion, and cod liver oil is an example of a water
in oil emulsion.
5. The given metal is a divalent ion = M
2+
. It has atomic number 25 and will have d
5
electronic configuration.
?The magnetic moment µ is
µ = 5 ( 5 + 2 ) = 5 . 9 2B M
OR
The order of increasing oxidising power is
7
2
2 2 4
VO Cr O MnO
? ? ?
??
Section B
6.
B
T 80.31 C 80.10 C 0.21 C ? ? ? ?
WB= 1.25 g, Kb =2.53 ?C kg mol
1-
, MB =?, WA = 99 g
bB
B
bA
B
K ×W
M = ×1000
?T ×W
2.53×1.25×1000
=
0.21×99
M =152g/mol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
OR
Given:
i = 3,
f
K = 1.86 K kg mol
-1
,
B
W = 10.5 g,
B
M = 184 g mol
-1
,
A
W = 200 g
fB
f
BA
i×K ×W ×1000
? T =
M ×W
On substituting the values in the above equation, we get
f
f
o
f f f
f
3×1.86×10.5×1000
? T = = 1 . 5 9 K
184×200
Aswe know ? T , l e t s f ind f r e e z ing point o f a q u e o ussolution,
T =T - ?T
T =273.15K-1.59K=271.56K
7. The resonating structures of NO2 are
The resonating structures of N2O2 are
8.
3 2 2 7 2 4
CaCO K Cr O /H SO
3 2 3 3 2 3 3 Oxidation
Ethanol Ethanoicacid Ca.Ethanoate Ethanol
CH CH OH CH COOH (CH COO) Ca CH COCH ? ? ? ? ? ? ? ? ? ? ? ? ?
9.
(aq )
(s) 2(s)
Cd 2OH Cd(OH) 2e
??
? ? ? ? ?
2(s) 2 2(s)
NiO +2H O 2e Ni(OH) +2OH
??
? ? ? ?
10.
(a) Molecularity: Pseudomolecular
(b) Order of reaction: First order (1)
11.
(a) When phenol is treated with FeCl3 solution, it gives violet colouration. Cyclohexanol
does not give any colouration.
3
6 5 3 6 5 6
Phenol Violet colouration
6C H OH FeCl [Fe(OC H ) ] 3H 3HCl
??
? ? ? ? ? ?
Cyclohexanol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
(b) In the iodoform test, isopropyl alcohol is warmed with NaOI (I2NaOH) to give a
yellow precipitate of idoform, while benzyl alcohol does not give any yellow
precipitate.
12. Addition polymer: polyvinyl chloride, polyethene
Condensation polymers: terylene, Bakelite
Section C
13.
(a)
(b)
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
(c)
14.
(a) Schottky defect
(b) Decreases
(c) Na
+
Cl
-
15.
(a) Distillation: Zn can vaporise easily, whereas Fe cannot. Therefore, they can be
separated by distillation.
(b) Liquation: Mg has a lower melting point than Cu. Therefore, molten magnesium will
slide down, whereas copper will be left behind at the top of the furnace.
(c) Rare earth metals can be separated by ion exchange methods due to a difference in
the size of their hydrated cations.
OR
Reactions taking place in different zones in the blast furnace during the extraction of
iron:
(a) At 500 –800 K
3 3 3 4 2
3 4 2
2 3 2
3Fe O CO 2Fe O CO
Fe O 4CO 3Fe 4CO
Fe O CO 2FeO CO
? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
(b) At 900 –1500 K
2
2
22
C CO 2CO
FeO CO Fe CO
C O CO
? ? ? ?
? ? ? ? ?
? ? ? ?
(c) Above 1570 K
32
23
Slag
FeO C Fe CO
CaCO CaO CO
CaO SiO CaSiO
?
? ? ? ? ?
? ? ? ?
? ? ? ?
Page 5
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
CBSE
Class XII Chemistry
Sample Paper 2 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. Due to intermolecular hydrogen bonding in alcohols, which is absent in ethers, alcohols
have a higher boiling point than ethers.
2. Polymerisation: It is the process in which a number of monomers join together to form
a long-chain polymer.
2
O ,350 570K
2 2 2 2 n 1000 2000atm
Ethene Polyethene
(monomer) (polymer)
nCH CH (CH CH )
?
?
? ? ? ? ? ? ? ? ? ?
3.
1
8(corneratom)× atomperunit cell=1
8
OR
When NaCl is doped with MgCl2, each Mg
2+
ion replaces 2 Na
+
ions to maintain electrical
neutrality. One site is occupied by Mg
2+
ion, while the other remains vacant. Thus, cation
vacancies are formed. This defect is called an impurity defect.
4. Milk is an example of an oil in water emulsion, and cod liver oil is an example of a water
in oil emulsion.
5. The given metal is a divalent ion = M
2+
. It has atomic number 25 and will have d
5
electronic configuration.
?The magnetic moment µ is
µ = 5 ( 5 + 2 ) = 5 . 9 2B M
OR
The order of increasing oxidising power is
7
2
2 2 4
VO Cr O MnO
? ? ?
??
Section B
6.
B
T 80.31 C 80.10 C 0.21 C ? ? ? ?
WB= 1.25 g, Kb =2.53 ?C kg mol
1-
, MB =?, WA = 99 g
bB
B
bA
B
K ×W
M = ×1000
?T ×W
2.53×1.25×1000
=
0.21×99
M =152g/mol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
OR
Given:
i = 3,
f
K = 1.86 K kg mol
-1
,
B
W = 10.5 g,
B
M = 184 g mol
-1
,
A
W = 200 g
fB
f
BA
i×K ×W ×1000
? T =
M ×W
On substituting the values in the above equation, we get
f
f
o
f f f
f
3×1.86×10.5×1000
? T = = 1 . 5 9 K
184×200
Aswe know ? T , l e t s f ind f r e e z ing point o f a q u e o ussolution,
T =T - ?T
T =273.15K-1.59K=271.56K
7. The resonating structures of NO2 are
The resonating structures of N2O2 are
8.
3 2 2 7 2 4
CaCO K Cr O /H SO
3 2 3 3 2 3 3 Oxidation
Ethanol Ethanoicacid Ca.Ethanoate Ethanol
CH CH OH CH COOH (CH COO) Ca CH COCH ? ? ? ? ? ? ? ? ? ? ? ? ?
9.
(aq )
(s) 2(s)
Cd 2OH Cd(OH) 2e
??
? ? ? ? ?
2(s) 2 2(s)
NiO +2H O 2e Ni(OH) +2OH
??
? ? ? ?
10.
(a) Molecularity: Pseudomolecular
(b) Order of reaction: First order (1)
11.
(a) When phenol is treated with FeCl3 solution, it gives violet colouration. Cyclohexanol
does not give any colouration.
3
6 5 3 6 5 6
Phenol Violet colouration
6C H OH FeCl [Fe(OC H ) ] 3H 3HCl
??
? ? ? ? ? ?
Cyclohexanol
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
(b) In the iodoform test, isopropyl alcohol is warmed with NaOI (I2NaOH) to give a
yellow precipitate of idoform, while benzyl alcohol does not give any yellow
precipitate.
12. Addition polymer: polyvinyl chloride, polyethene
Condensation polymers: terylene, Bakelite
Section C
13.
(a)
(b)
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
(c)
14.
(a) Schottky defect
(b) Decreases
(c) Na
+
Cl
-
15.
(a) Distillation: Zn can vaporise easily, whereas Fe cannot. Therefore, they can be
separated by distillation.
(b) Liquation: Mg has a lower melting point than Cu. Therefore, molten magnesium will
slide down, whereas copper will be left behind at the top of the furnace.
(c) Rare earth metals can be separated by ion exchange methods due to a difference in
the size of their hydrated cations.
OR
Reactions taking place in different zones in the blast furnace during the extraction of
iron:
(a) At 500 –800 K
3 3 3 4 2
3 4 2
2 3 2
3Fe O CO 2Fe O CO
Fe O 4CO 3Fe 4CO
Fe O CO 2FeO CO
? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
(b) At 900 –1500 K
2
2
22
C CO 2CO
FeO CO Fe CO
C O CO
? ? ? ?
? ? ? ? ?
? ? ? ?
(c) Above 1570 K
32
23
Slag
FeO C Fe CO
CaCO CaO CO
CaO SiO CaSiO
?
? ? ? ? ?
? ? ? ?
? ? ? ?
CBSE XII | CHEMISTRY
Sample Paper 2 - Solution
16. Given:
Moles of glucose = 18 g/180 g/mol = 0.1 mol
Mass of the solvent = 1 kg
Hence, molarity of glucose solution =
1
0.1 mol
0.1molkg
1Kg
?
?
Change in boiling point will be
bb
b
? T = K ×m
? T = 0 . 5 2×0.1=0.052K
Water boils at 373.15 K at 1.013 bar pressure.
So, the boiling point will be
373.15 + 0.052 = 373.202 K
OR
The boiling point is the temperature at which the vapour pressure of the solution is
equal to the atmospheric pressure of the solution. In a solution containing a non-
volatile solute, the vapour pressure is lowered as the surface of the liquid is also
occupied by some molecules of a non-volatile solute.
Hence, high temperature is required to make the vapour pressure of the solution equal
to the atmospheric pressure which eventually increases the boiling point of the
solution.
17.
(a)
Step 1:
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