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 Page 1


   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
CBSE 
Class XII Chemistry  
Sample Paper 1 - Solution 
 
 
Time: 3 Hrs                                                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 Section A 
1. A = 
1
8 × = 1
8
 
             B =  
1
6 × = 3
2
  
           Formula = AB3 
OR 
 
         Dispersion forces, dipole-dipole interactions and hydrogen bonds. 
 
2.  Dispersed phase: Liquid  
        Dispersion medium: Gas 
 
3. +3 is the most common oxidation state of lanthanoids.  
                
OR 
 
The most common oxidation state of Cu, Ag and Au is +1. 
 
4. 1-bromo-2, 2-dimethyl propane   
                             
5. 
2 2 2 2 5
Glycine Amino caproic acid
H N-CH -COOH and H N-(CH ) COOH  
        
 Section B 
 
6. Henry’s law: It states that ‘the partial pressure of the gas in vapour phase (p) is 
proportional to the mole fraction of the gas (x) in the solution’ and expressed as: 
 
P = KHx 
     
where KH is Henry’s constant. 
 
Significance of KH: Higher the value of Henry’s law constant KH, lower is the solubility of the 
gas in the liquid. 
 
OR 
 
In dilute solutions, 
+ 2-
2 4 4
K SO 2K + SO ? ? ? 
No.of molesof particlesafterdissociation 3
van't Hoff Factor,i 3
No.of molesof particlesbeforedissociation 1
? ? ? 
Page 2


   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
CBSE 
Class XII Chemistry  
Sample Paper 1 - Solution 
 
 
Time: 3 Hrs                                                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 Section A 
1. A = 
1
8 × = 1
8
 
             B =  
1
6 × = 3
2
  
           Formula = AB3 
OR 
 
         Dispersion forces, dipole-dipole interactions and hydrogen bonds. 
 
2.  Dispersed phase: Liquid  
        Dispersion medium: Gas 
 
3. +3 is the most common oxidation state of lanthanoids.  
                
OR 
 
The most common oxidation state of Cu, Ag and Au is +1. 
 
4. 1-bromo-2, 2-dimethyl propane   
                             
5. 
2 2 2 2 5
Glycine Amino caproic acid
H N-CH -COOH and H N-(CH ) COOH  
        
 Section B 
 
6. Henry’s law: It states that ‘the partial pressure of the gas in vapour phase (p) is 
proportional to the mole fraction of the gas (x) in the solution’ and expressed as: 
 
P = KHx 
     
where KH is Henry’s constant. 
 
Significance of KH: Higher the value of Henry’s law constant KH, lower is the solubility of the 
gas in the liquid. 
 
OR 
 
In dilute solutions, 
+ 2-
2 4 4
K SO 2K + SO ? ? ? 
No.of molesof particlesafterdissociation 3
van't Hoff Factor,i 3
No.of molesof particlesbeforedissociation 1
? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
7.  
(a) Alkalinity of the solution prevents the availability of H
+ 
ions. 
(b) A galvanised substance is that which has been coated with a layer of zinc to delay 
corrosion. Zinc gets corroded instead of the substance.  
The outer layer of zinc of any galvanised material reacts with the atmospheric 
oxygen to form zinc oxide (ZnO), which is stronger than zinc. Thus, even if the outer 
layer of zinc undergoes corrosion, the material is getting coated with a stronger 
substance (ZnO), and thus is better able to resist corrosion. 
 
8.  
(a) 
13
Order= + =2, i.e., second order
22
 
(b) 
31
Order= +(-1) = , i.e., half order
22
 
 
9.   
(a) 2XeF2 + 2H2O    ?     2Xe + 4HF + O2 
(b) 2PH3 + 3HgCl2    ?    Hg3P2  +  6HCl 
   
10.  
(a) When phenol reacts with Br2 in CS2 at 273 K, a mixture of o- and p-bromophenol is 
formed in which p-bromophenol is the major product.                                                                      
                                                                                                 
(b) When phenol reacts with conc. HNO3, 2,4,6-trinitrophenol is formed.     
                     
11.  Acetophenone (C6H5COCH3) contains the group (–CH3CO) and hence gives the iodoform 
test, while benzophenone does not contain this group and hence does not give the 
iodoform test: 
6 5 3 2 3 6 5 2
Acetophenone Idoform
C H COCH +3I +4NaOH CHI + C H COONa + 3NaI + 3H O ? ? ? 
 
6 5 6 5 2
Benzophenone
C H COC H +3I +4NaOH No Reaction ? ? ? 
Page 3


   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
CBSE 
Class XII Chemistry  
Sample Paper 1 - Solution 
 
 
Time: 3 Hrs                                                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 Section A 
1. A = 
1
8 × = 1
8
 
             B =  
1
6 × = 3
2
  
           Formula = AB3 
OR 
 
         Dispersion forces, dipole-dipole interactions and hydrogen bonds. 
 
2.  Dispersed phase: Liquid  
        Dispersion medium: Gas 
 
3. +3 is the most common oxidation state of lanthanoids.  
                
OR 
 
The most common oxidation state of Cu, Ag and Au is +1. 
 
4. 1-bromo-2, 2-dimethyl propane   
                             
5. 
2 2 2 2 5
Glycine Amino caproic acid
H N-CH -COOH and H N-(CH ) COOH  
        
 Section B 
 
6. Henry’s law: It states that ‘the partial pressure of the gas in vapour phase (p) is 
proportional to the mole fraction of the gas (x) in the solution’ and expressed as: 
 
P = KHx 
     
where KH is Henry’s constant. 
 
Significance of KH: Higher the value of Henry’s law constant KH, lower is the solubility of the 
gas in the liquid. 
 
OR 
 
In dilute solutions, 
+ 2-
2 4 4
K SO 2K + SO ? ? ? 
No.of molesof particlesafterdissociation 3
van't Hoff Factor,i 3
No.of molesof particlesbeforedissociation 1
? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
7.  
(a) Alkalinity of the solution prevents the availability of H
+ 
ions. 
(b) A galvanised substance is that which has been coated with a layer of zinc to delay 
corrosion. Zinc gets corroded instead of the substance.  
The outer layer of zinc of any galvanised material reacts with the atmospheric 
oxygen to form zinc oxide (ZnO), which is stronger than zinc. Thus, even if the outer 
layer of zinc undergoes corrosion, the material is getting coated with a stronger 
substance (ZnO), and thus is better able to resist corrosion. 
 
8.  
(a) 
13
Order= + =2, i.e., second order
22
 
(b) 
31
Order= +(-1) = , i.e., half order
22
 
 
9.   
(a) 2XeF2 + 2H2O    ?     2Xe + 4HF + O2 
(b) 2PH3 + 3HgCl2    ?    Hg3P2  +  6HCl 
   
10.  
(a) When phenol reacts with Br2 in CS2 at 273 K, a mixture of o- and p-bromophenol is 
formed in which p-bromophenol is the major product.                                                                      
                                                                                                 
(b) When phenol reacts with conc. HNO3, 2,4,6-trinitrophenol is formed.     
                     
11.  Acetophenone (C6H5COCH3) contains the group (–CH3CO) and hence gives the iodoform 
test, while benzophenone does not contain this group and hence does not give the 
iodoform test: 
6 5 3 2 3 6 5 2
Acetophenone Idoform
C H COCH +3I +4NaOH CHI + C H COONa + 3NaI + 3H O ? ? ? 
 
6 5 6 5 2
Benzophenone
C H COC H +3I +4NaOH No Reaction ? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
OR 
(a)  
 
(b)  
 
12.    
Polymer Monomer Structure of the monomer 
(a) PVC Vinyl chloride CH 2=CH=Cl 
(b) Polypropene Propene CH 3-CH=CH 2 
 
Section C 
 
13.      Diameter = 245 pm 
             
245
Radius pm 122.5pm
2
3
Inabccstructure,r a
4
3
122.5 a
4
122.5 4 490
a 282.91pm
1.732
3
? ? ?
?
??
?
? ? ? ?
 
 
                
3 -10 3 23
A
-3
-23 23
z×M 2×52
d= =
a ×N (282.91×10 ) ×6.02×10
104 104
= = =7.63gcm
2.264×10 ×6.02×10 2.264×6.02
                
                                                           
14.  According to Rault’s law, 
         
A B A
A B A
P P W M
P M W
?
?? 
Here, 
A
P = 3.64 kPa, p = 2.8 kPa, WB = 30 g, MA = 18 g mol
-1
, MB = ?, WA = 90 g 
Substituting the values in the above equation, we get 
        
-1
B
-1
-1
B
3.64kPa-2.80kPa 30g×18gmol
=
3.64kPa M ×90g
30g×18gmol ×3.64kPa
M = =26gmol
90g×0.84KPa
 
 
Page 4


   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
CBSE 
Class XII Chemistry  
Sample Paper 1 - Solution 
 
 
Time: 3 Hrs                                                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 Section A 
1. A = 
1
8 × = 1
8
 
             B =  
1
6 × = 3
2
  
           Formula = AB3 
OR 
 
         Dispersion forces, dipole-dipole interactions and hydrogen bonds. 
 
2.  Dispersed phase: Liquid  
        Dispersion medium: Gas 
 
3. +3 is the most common oxidation state of lanthanoids.  
                
OR 
 
The most common oxidation state of Cu, Ag and Au is +1. 
 
4. 1-bromo-2, 2-dimethyl propane   
                             
5. 
2 2 2 2 5
Glycine Amino caproic acid
H N-CH -COOH and H N-(CH ) COOH  
        
 Section B 
 
6. Henry’s law: It states that ‘the partial pressure of the gas in vapour phase (p) is 
proportional to the mole fraction of the gas (x) in the solution’ and expressed as: 
 
P = KHx 
     
where KH is Henry’s constant. 
 
Significance of KH: Higher the value of Henry’s law constant KH, lower is the solubility of the 
gas in the liquid. 
 
OR 
 
In dilute solutions, 
+ 2-
2 4 4
K SO 2K + SO ? ? ? 
No.of molesof particlesafterdissociation 3
van't Hoff Factor,i 3
No.of molesof particlesbeforedissociation 1
? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
7.  
(a) Alkalinity of the solution prevents the availability of H
+ 
ions. 
(b) A galvanised substance is that which has been coated with a layer of zinc to delay 
corrosion. Zinc gets corroded instead of the substance.  
The outer layer of zinc of any galvanised material reacts with the atmospheric 
oxygen to form zinc oxide (ZnO), which is stronger than zinc. Thus, even if the outer 
layer of zinc undergoes corrosion, the material is getting coated with a stronger 
substance (ZnO), and thus is better able to resist corrosion. 
 
8.  
(a) 
13
Order= + =2, i.e., second order
22
 
(b) 
31
Order= +(-1) = , i.e., half order
22
 
 
9.   
(a) 2XeF2 + 2H2O    ?     2Xe + 4HF + O2 
(b) 2PH3 + 3HgCl2    ?    Hg3P2  +  6HCl 
   
10.  
(a) When phenol reacts with Br2 in CS2 at 273 K, a mixture of o- and p-bromophenol is 
formed in which p-bromophenol is the major product.                                                                      
                                                                                                 
(b) When phenol reacts with conc. HNO3, 2,4,6-trinitrophenol is formed.     
                     
11.  Acetophenone (C6H5COCH3) contains the group (–CH3CO) and hence gives the iodoform 
test, while benzophenone does not contain this group and hence does not give the 
iodoform test: 
6 5 3 2 3 6 5 2
Acetophenone Idoform
C H COCH +3I +4NaOH CHI + C H COONa + 3NaI + 3H O ? ? ? 
 
6 5 6 5 2
Benzophenone
C H COC H +3I +4NaOH No Reaction ? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
OR 
(a)  
 
(b)  
 
12.    
Polymer Monomer Structure of the monomer 
(a) PVC Vinyl chloride CH 2=CH=Cl 
(b) Polypropene Propene CH 3-CH=CH 2 
 
Section C 
 
13.      Diameter = 245 pm 
             
245
Radius pm 122.5pm
2
3
Inabccstructure,r a
4
3
122.5 a
4
122.5 4 490
a 282.91pm
1.732
3
? ? ?
?
??
?
? ? ? ?
 
 
                
3 -10 3 23
A
-3
-23 23
z×M 2×52
d= =
a ×N (282.91×10 ) ×6.02×10
104 104
= = =7.63gcm
2.264×10 ×6.02×10 2.264×6.02
                
                                                           
14.  According to Rault’s law, 
         
A B A
A B A
P P W M
P M W
?
?? 
Here, 
A
P = 3.64 kPa, p = 2.8 kPa, WB = 30 g, MA = 18 g mol
-1
, MB = ?, WA = 90 g 
Substituting the values in the above equation, we get 
        
-1
B
-1
-1
B
3.64kPa-2.80kPa 30g×18gmol
=
3.64kPa M ×90g
30g×18gmol ×3.64kPa
M = =26gmol
90g×0.84KPa
 
 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
OR 
 
       
f
ff
11
f
f
f
f
T [273.15 ( 0.302 273.15)]K 0.320K
T K .m
T 1.86K kg mol 0.0711molKg
T 0.132K
Observed value of T 0.320K
i 2.42
alculated value of T 0.132K
??
? ? ? ? ? ?
??
? ? ?
??
?
? ? ?
?
 
 
 
15.  
(a) For a first-order reaction,     
   
2.303 a
k = log     .........(1)  
t a - x
          
  
Here, a = 0.062 atm 
     a - x = 0.044 atm
         t =  55 s
 
 
  
-3 -1
2.303 0.062
k =  log = 6.24x10 s
55 0.044
 
 
(b)  To calculate a - x when t = 100 s, a = 0.062 atm       
Putting the values in the above equation, 
 
         
?
-3
2.303 0.062
6.24x10 = log
100 a - x
  
                  ?a - x = 0.033 atm     
                                        
 
16.  
(a) River water is muddy and contains charged colloidal particles of clay, sand and 
many other materials. Sea water contains in it a number of dissolved electrolytes. 
When sea water and river water come in contact with each other, the electrolytes 
present in sea water coagulate the suspended colloidal particles which ultimately 
settle at the point of contact. Thus, a delta is formed at the point where the river 
enters the sea. 
 
(b) Characteristics of catalysts:  
(i) Catalysts are highly selective. A catalyst is able to direct a reaction to give a 
particular product.    
(ii) Catalysts are highly active. A catalyst is able to increase the rate of a chemical 
reaction.  
 
(c) A colloid in which the particles of dispersed phase are sufficiently big in size to be of 
colloidal dimensions is called a macromolecular colloid. Example: Starch  
Page 5


   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
CBSE 
Class XII Chemistry  
Sample Paper 1 - Solution 
 
 
Time: 3 Hrs                                                                              Total Marks: 70 
_________________________________________________________________________________________________________ 
 Section A 
1. A = 
1
8 × = 1
8
 
             B =  
1
6 × = 3
2
  
           Formula = AB3 
OR 
 
         Dispersion forces, dipole-dipole interactions and hydrogen bonds. 
 
2.  Dispersed phase: Liquid  
        Dispersion medium: Gas 
 
3. +3 is the most common oxidation state of lanthanoids.  
                
OR 
 
The most common oxidation state of Cu, Ag and Au is +1. 
 
4. 1-bromo-2, 2-dimethyl propane   
                             
5. 
2 2 2 2 5
Glycine Amino caproic acid
H N-CH -COOH and H N-(CH ) COOH  
        
 Section B 
 
6. Henry’s law: It states that ‘the partial pressure of the gas in vapour phase (p) is 
proportional to the mole fraction of the gas (x) in the solution’ and expressed as: 
 
P = KHx 
     
where KH is Henry’s constant. 
 
Significance of KH: Higher the value of Henry’s law constant KH, lower is the solubility of the 
gas in the liquid. 
 
OR 
 
In dilute solutions, 
+ 2-
2 4 4
K SO 2K + SO ? ? ? 
No.of molesof particlesafterdissociation 3
van't Hoff Factor,i 3
No.of molesof particlesbeforedissociation 1
? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
7.  
(a) Alkalinity of the solution prevents the availability of H
+ 
ions. 
(b) A galvanised substance is that which has been coated with a layer of zinc to delay 
corrosion. Zinc gets corroded instead of the substance.  
The outer layer of zinc of any galvanised material reacts with the atmospheric 
oxygen to form zinc oxide (ZnO), which is stronger than zinc. Thus, even if the outer 
layer of zinc undergoes corrosion, the material is getting coated with a stronger 
substance (ZnO), and thus is better able to resist corrosion. 
 
8.  
(a) 
13
Order= + =2, i.e., second order
22
 
(b) 
31
Order= +(-1) = , i.e., half order
22
 
 
9.   
(a) 2XeF2 + 2H2O    ?     2Xe + 4HF + O2 
(b) 2PH3 + 3HgCl2    ?    Hg3P2  +  6HCl 
   
10.  
(a) When phenol reacts with Br2 in CS2 at 273 K, a mixture of o- and p-bromophenol is 
formed in which p-bromophenol is the major product.                                                                      
                                                                                                 
(b) When phenol reacts with conc. HNO3, 2,4,6-trinitrophenol is formed.     
                     
11.  Acetophenone (C6H5COCH3) contains the group (–CH3CO) and hence gives the iodoform 
test, while benzophenone does not contain this group and hence does not give the 
iodoform test: 
6 5 3 2 3 6 5 2
Acetophenone Idoform
C H COCH +3I +4NaOH CHI + C H COONa + 3NaI + 3H O ? ? ? 
 
6 5 6 5 2
Benzophenone
C H COC H +3I +4NaOH No Reaction ? ? ? 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
OR 
(a)  
 
(b)  
 
12.    
Polymer Monomer Structure of the monomer 
(a) PVC Vinyl chloride CH 2=CH=Cl 
(b) Polypropene Propene CH 3-CH=CH 2 
 
Section C 
 
13.      Diameter = 245 pm 
             
245
Radius pm 122.5pm
2
3
Inabccstructure,r a
4
3
122.5 a
4
122.5 4 490
a 282.91pm
1.732
3
? ? ?
?
??
?
? ? ? ?
 
 
                
3 -10 3 23
A
-3
-23 23
z×M 2×52
d= =
a ×N (282.91×10 ) ×6.02×10
104 104
= = =7.63gcm
2.264×10 ×6.02×10 2.264×6.02
                
                                                           
14.  According to Rault’s law, 
         
A B A
A B A
P P W M
P M W
?
?? 
Here, 
A
P = 3.64 kPa, p = 2.8 kPa, WB = 30 g, MA = 18 g mol
-1
, MB = ?, WA = 90 g 
Substituting the values in the above equation, we get 
        
-1
B
-1
-1
B
3.64kPa-2.80kPa 30g×18gmol
=
3.64kPa M ×90g
30g×18gmol ×3.64kPa
M = =26gmol
90g×0.84KPa
 
 
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
OR 
 
       
f
ff
11
f
f
f
f
T [273.15 ( 0.302 273.15)]K 0.320K
T K .m
T 1.86K kg mol 0.0711molKg
T 0.132K
Observed value of T 0.320K
i 2.42
alculated value of T 0.132K
??
? ? ? ? ? ?
??
? ? ?
??
?
? ? ?
?
 
 
 
15.  
(a) For a first-order reaction,     
   
2.303 a
k = log     .........(1)  
t a - x
          
  
Here, a = 0.062 atm 
     a - x = 0.044 atm
         t =  55 s
 
 
  
-3 -1
2.303 0.062
k =  log = 6.24x10 s
55 0.044
 
 
(b)  To calculate a - x when t = 100 s, a = 0.062 atm       
Putting the values in the above equation, 
 
         
?
-3
2.303 0.062
6.24x10 = log
100 a - x
  
                  ?a - x = 0.033 atm     
                                        
 
16.  
(a) River water is muddy and contains charged colloidal particles of clay, sand and 
many other materials. Sea water contains in it a number of dissolved electrolytes. 
When sea water and river water come in contact with each other, the electrolytes 
present in sea water coagulate the suspended colloidal particles which ultimately 
settle at the point of contact. Thus, a delta is formed at the point where the river 
enters the sea. 
 
(b) Characteristics of catalysts:  
(i) Catalysts are highly selective. A catalyst is able to direct a reaction to give a 
particular product.    
(ii) Catalysts are highly active. A catalyst is able to increase the rate of a chemical 
reaction.  
 
(c) A colloid in which the particles of dispersed phase are sufficiently big in size to be of 
colloidal dimensions is called a macromolecular colloid. Example: Starch  
   
 
CBSE XII  |  Chemistry  
Sample Paper 1 - Solution 
 
     
17.  
(a) The principle of the froth flotation process is that sulphide ore particles are 
preferentially wetted by oil, whereas gangue particles are wetted by water. 
 
(b) Zone refining is based on the principle that the impurities are more soluble in the 
melt than in the solid state of the metal. 
 
(c) The principle of refining by liquation is that the impurities whose melting points are 
higher than the metal are left behind on melting the impure metal. Hence, pure 
metal separates out. 
 
18.  
(a) 
2
42
2
2 4 2
22
4 2 4 2 2
MnO 8H 5e Mn 4H O] 2
C O 2CO 2e ] 5
________________________________________________________
2MnO 5C O 16H 2Mn 10CO 8H O
? ? ? ?
??
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ? ?
 
 
 
(b) 
Heat
4 2 4 2 2
2KMnO K MnO MnO O ? ? ? ? ? 
 
(c)  
23
2 7 2
2
23
2 7 2 2
Cr O 14H 6e 2Cr 7H O
H S 2H S 2e ] 3
_________________________________________________
Cr O H S 14H 2Cr 3S 7H O
? ? ? ?
??
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
  
 
 
19.  
(a) Potassium trioxalatochromate (III) 
 
(b) [CoF6]
3-
 
Co (Z=27): [Ar] 3d
7
4s
2
 
Co
3+
: [Ar] 3d
6 
4s
0 
F
-
 is a weak field ligand and therefore does not cause pairing of electrons. 
There are four unpaired electrons. 
[Co(NH3)6]
3+ 
Co (Z = 27): [Ar]3d
7
4s
2
 
Co
3+
: [Ar]3d
6
4s
0 
NH3 is a strong field ligand and causes pairing of electrons. Hence, there are no 
unpaired electrons.     
                                                    
(c) Hydrate isomerism     
 
 
 
 
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