Sample Solution Paper 2 - Chemistry, Class 12 NEET Notes | EduRev

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 Page 1


  
 
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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