Chemistry: Sample Solution Paper- 5, Class 12 Notes | EduRev

 Page 1


  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
CBSE  
Class  XII Chemistry   
Sample  Paper 5 - Solution 
 
 Time: 3 Hrs                                                                Total Marks: 70 
 
Section A 
 
1. Amorphous solid                                                                                                                              
OR 
 
The coordination number of hcp and ccp is 12.  
 
2. A finely divided substance provides more surface area for adsorption and hence is more 
effective as an absorbent.                                                                                                                   
 
 
3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for 
lighters. 
OR 
 
KMnO4 has +7 oxidation state. 
 
4. Cyclopentane carbaldehyde 
 
5. Phenol and formaldehyde 
 
Section B 
6. Molality   
                    
222.6
= = 17.95m
62× 0.2
    
 
          Mass of solution = 200 + 222.6 = 422.6 g 
          
-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL
   
                                                
           Molarity 
B
B
w 1000
=×
M Volumeof solution(in mL)
   
                             
222.6 1000
=×
62 394.22
=9.1M
  
  
 
OR 
 
Page 2


  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
CBSE  
Class  XII Chemistry   
Sample  Paper 5 - Solution 
 
 Time: 3 Hrs                                                                Total Marks: 70 
 
Section A 
 
1. Amorphous solid                                                                                                                              
OR 
 
The coordination number of hcp and ccp is 12.  
 
2. A finely divided substance provides more surface area for adsorption and hence is more 
effective as an absorbent.                                                                                                                   
 
 
3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for 
lighters. 
OR 
 
KMnO4 has +7 oxidation state. 
 
4. Cyclopentane carbaldehyde 
 
5. Phenol and formaldehyde 
 
Section B 
6. Molality   
                    
222.6
= = 17.95m
62× 0.2
    
 
          Mass of solution = 200 + 222.6 = 422.6 g 
          
-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL
   
                                                
           Molarity 
B
B
w 1000
=×
M Volumeof solution(in mL)
   
                             
222.6 1000
=×
62 394.22
=9.1M
  
  
 
OR 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
       The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g 
mol
-1
. 
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?
 
                    
 
7.  
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??
 
 
8.  
        (a) 
1
s
?
                                                                                                
        (b) Slope = -k           
                                                                            
9.  
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-
 
ion is known, but phosphorus in CP
-
 ion does not form ???? - ???? bond due to its 
bigger size. 
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore 
gets dimerised to the stable N2O4. 
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2. 
 
10.  
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to 
form anilinium ion, which is a meta directing group, thus giving substantial amount 
of m-nitroaniline.   
 
(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene 
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its 
pKb value will be higher. On the other hand, the +I effect of the 
3
-CH group, 
increases the electron density on the N atom in 
32
CH NH making it a stronger base. 
Hence, its Kb value will be higher and its pKb value will be lower.                                                                                                                  
 
OR 
Page 3


  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
CBSE  
Class  XII Chemistry   
Sample  Paper 5 - Solution 
 
 Time: 3 Hrs                                                                Total Marks: 70 
 
Section A 
 
1. Amorphous solid                                                                                                                              
OR 
 
The coordination number of hcp and ccp is 12.  
 
2. A finely divided substance provides more surface area for adsorption and hence is more 
effective as an absorbent.                                                                                                                   
 
 
3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for 
lighters. 
OR 
 
KMnO4 has +7 oxidation state. 
 
4. Cyclopentane carbaldehyde 
 
5. Phenol and formaldehyde 
 
Section B 
6. Molality   
                    
222.6
= = 17.95m
62× 0.2
    
 
          Mass of solution = 200 + 222.6 = 422.6 g 
          
-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL
   
                                                
           Molarity 
B
B
w 1000
=×
M Volumeof solution(in mL)
   
                             
222.6 1000
=×
62 394.22
=9.1M
  
  
 
OR 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
       The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g 
mol
-1
. 
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?
 
                    
 
7.  
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??
 
 
8.  
        (a) 
1
s
?
                                                                                                
        (b) Slope = -k           
                                                                            
9.  
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-
 
ion is known, but phosphorus in CP
-
 ion does not form ???? - ???? bond due to its 
bigger size. 
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore 
gets dimerised to the stable N2O4. 
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2. 
 
10.  
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to 
form anilinium ion, which is a meta directing group, thus giving substantial amount 
of m-nitroaniline.   
 
(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene 
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its 
pKb value will be higher. On the other hand, the +I effect of the 
3
-CH group, 
increases the electron density on the N atom in 
32
CH NH making it a stronger base. 
Hence, its Kb value will be higher and its pKb value will be lower.                                                                                                                  
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
 
(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom 
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen 
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding. 
Thus, primary amines have a higher boiling point than tertiary amines as a result of 
their hydrogen bonding. 
 
(b)   
                           +                           
                                                                               
 
11.  
(a) Butanone < Propanone < Propanal < Ethanal 
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde 
 
 
12.  
(a) Polymers which are degraded by microorganisms within a suitable period so that 
the polymers and their degraded products do not cause any serious effects on the 
environment are called biodegradable polymers.                           
              Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)                      
 
(b) On the basis of molecular forces present between the chains of various polymers, 
polymers are classified as 
 1. Elastomers 
  2. Fibres 
3. Thermoplastics 
4. Thermosetting plastics 
 
Section C 
 
13.  
(a) The co-ordination number of each sphere in the bcc structure is 8.                                                                                             
(b) Number of A atoms = 
1
8x
8
 = 1 atom  
       Number of B atoms = 
1
6x
2
 = 3 atoms 
Therefore, the formula of the crystalline compound = AB3 
(c) The compound of group 12–16 is zinc sulphide.  
 
Page 4


  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
CBSE  
Class  XII Chemistry   
Sample  Paper 5 - Solution 
 
 Time: 3 Hrs                                                                Total Marks: 70 
 
Section A 
 
1. Amorphous solid                                                                                                                              
OR 
 
The coordination number of hcp and ccp is 12.  
 
2. A finely divided substance provides more surface area for adsorption and hence is more 
effective as an absorbent.                                                                                                                   
 
 
3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for 
lighters. 
OR 
 
KMnO4 has +7 oxidation state. 
 
4. Cyclopentane carbaldehyde 
 
5. Phenol and formaldehyde 
 
Section B 
6. Molality   
                    
222.6
= = 17.95m
62× 0.2
    
 
          Mass of solution = 200 + 222.6 = 422.6 g 
          
-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL
   
                                                
           Molarity 
B
B
w 1000
=×
M Volumeof solution(in mL)
   
                             
222.6 1000
=×
62 394.22
=9.1M
  
  
 
OR 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
       The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g 
mol
-1
. 
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?
 
                    
 
7.  
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??
 
 
8.  
        (a) 
1
s
?
                                                                                                
        (b) Slope = -k           
                                                                            
9.  
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-
 
ion is known, but phosphorus in CP
-
 ion does not form ???? - ???? bond due to its 
bigger size. 
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore 
gets dimerised to the stable N2O4. 
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2. 
 
10.  
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to 
form anilinium ion, which is a meta directing group, thus giving substantial amount 
of m-nitroaniline.   
 
(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene 
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its 
pKb value will be higher. On the other hand, the +I effect of the 
3
-CH group, 
increases the electron density on the N atom in 
32
CH NH making it a stronger base. 
Hence, its Kb value will be higher and its pKb value will be lower.                                                                                                                  
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
 
(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom 
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen 
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding. 
Thus, primary amines have a higher boiling point than tertiary amines as a result of 
their hydrogen bonding. 
 
(b)   
                           +                           
                                                                               
 
11.  
(a) Butanone < Propanone < Propanal < Ethanal 
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde 
 
 
12.  
(a) Polymers which are degraded by microorganisms within a suitable period so that 
the polymers and their degraded products do not cause any serious effects on the 
environment are called biodegradable polymers.                           
              Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)                      
 
(b) On the basis of molecular forces present between the chains of various polymers, 
polymers are classified as 
 1. Elastomers 
  2. Fibres 
3. Thermoplastics 
4. Thermosetting plastics 
 
Section C 
 
13.  
(a) The co-ordination number of each sphere in the bcc structure is 8.                                                                                             
(b) Number of A atoms = 
1
8x
8
 = 1 atom  
       Number of B atoms = 
1
6x
2
 = 3 atoms 
Therefore, the formula of the crystalline compound = AB3 
(c) The compound of group 12–16 is zinc sulphide.  
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
14.  
    ?
ff
B
f
BA
fB
B
fA
?T = K m
w 1000
= K ×
Mw
K × 1000 x w
M=
?T ×w
4.9 × 1000 × 2
=
1.62 × 25
-1
= 241.98 g mol
  
          
                                                                                
                    Normal molar mass of benzoic acid = 122 g mol
-1 
               i = 
NormalMolarMass 122
= =0.504
ObservedMolar Mass 241.98
    
 
             
? ? ? 2 C H COOH (C H COOH)
6 6 2 55
Initial 1 0
After
a
association 1 - a
2
Therefore,
a
1- a+
2
i=
1
a
i=1 -
2
a
i=1 -
2
2i=2- a
a = 2 - 2i
a = 2 - 2 x 0. 50 4
=2 -1.008
=0.992
a = 99 .2 %
      
 
 
 
 
 
 
 
Page 5


  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
CBSE  
Class  XII Chemistry   
Sample  Paper 5 - Solution 
 
 Time: 3 Hrs                                                                Total Marks: 70 
 
Section A 
 
1. Amorphous solid                                                                                                                              
OR 
 
The coordination number of hcp and ccp is 12.  
 
2. A finely divided substance provides more surface area for adsorption and hence is more 
effective as an absorbent.                                                                                                                   
 
 
3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for 
lighters. 
OR 
 
KMnO4 has +7 oxidation state. 
 
4. Cyclopentane carbaldehyde 
 
5. Phenol and formaldehyde 
 
Section B 
6. Molality   
                    
222.6
= = 17.95m
62× 0.2
    
 
          Mass of solution = 200 + 222.6 = 422.6 g 
          
-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL
   
                                                
           Molarity 
B
B
w 1000
=×
M Volumeof solution(in mL)
   
                             
222.6 1000
=×
62 394.22
=9.1M
  
  
 
OR 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
       The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g 
mol
-1
. 
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?
 
                    
 
7.  
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??
 
 
8.  
        (a) 
1
s
?
                                                                                                
        (b) Slope = -k           
                                                                            
9.  
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-
 
ion is known, but phosphorus in CP
-
 ion does not form ???? - ???? bond due to its 
bigger size. 
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore 
gets dimerised to the stable N2O4. 
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2. 
 
10.  
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to 
form anilinium ion, which is a meta directing group, thus giving substantial amount 
of m-nitroaniline.   
 
(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene 
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its 
pKb value will be higher. On the other hand, the +I effect of the 
3
-CH group, 
increases the electron density on the N atom in 
32
CH NH making it a stronger base. 
Hence, its Kb value will be higher and its pKb value will be lower.                                                                                                                  
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
 
(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom 
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen 
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding. 
Thus, primary amines have a higher boiling point than tertiary amines as a result of 
their hydrogen bonding. 
 
(b)   
                           +                           
                                                                               
 
11.  
(a) Butanone < Propanone < Propanal < Ethanal 
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde 
 
 
12.  
(a) Polymers which are degraded by microorganisms within a suitable period so that 
the polymers and their degraded products do not cause any serious effects on the 
environment are called biodegradable polymers.                           
              Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)                      
 
(b) On the basis of molecular forces present between the chains of various polymers, 
polymers are classified as 
 1. Elastomers 
  2. Fibres 
3. Thermoplastics 
4. Thermosetting plastics 
 
Section C 
 
13.  
(a) The co-ordination number of each sphere in the bcc structure is 8.                                                                                             
(b) Number of A atoms = 
1
8x
8
 = 1 atom  
       Number of B atoms = 
1
6x
2
 = 3 atoms 
Therefore, the formula of the crystalline compound = AB3 
(c) The compound of group 12–16 is zinc sulphide.  
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
14.  
    ?
ff
B
f
BA
fB
B
fA
?T = K m
w 1000
= K ×
Mw
K × 1000 x w
M=
?T ×w
4.9 × 1000 × 2
=
1.62 × 25
-1
= 241.98 g mol
  
          
                                                                                
                    Normal molar mass of benzoic acid = 122 g mol
-1 
               i = 
NormalMolarMass 122
= =0.504
ObservedMolar Mass 241.98
    
 
             
? ? ? 2 C H COOH (C H COOH)
6 6 2 55
Initial 1 0
After
a
association 1 - a
2
Therefore,
a
1- a+
2
i=
1
a
i=1 -
2
a
i=1 -
2
2i=2- a
a = 2 - 2i
a = 2 - 2 x 0. 50 4
=2 -1.008
=0.992
a = 99 .2 %
      
 
 
 
 
 
 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  5 - Solution 
 
     
15.    For a first-order reaction: 
  
2.303 a
k = log
t a - x
       
       
-5
2.303 a
2.3 × 10 = log
200 × 60 a - x
a
log =0.1198
a - x
a
= Antilog (0.1198)
a - x
a
=1.317
a - x
a = 1.317(a - x)
1 a - x
=
1.317 a
x
1 - =0.759
a
x
=1-0.759=0.241
a
   
  
        
x
= 0.241
a
                                                                      
 % decomposed = 24.1%  
   % remained = 100 - 24.1 = 75.9%  
OR 
 
 
 
16.  
(a) Alcosol: A colloidal sol in which the dispersion medium is alcohol. Example: 
Collodion 
(b) Aerosol: When the dispersion medium is a gas and the dispersed phase is either 
solid or liquid, the colloidal system is called an aerosol. Examples: Fog, cloud, smoke 
(c) Hydrosol: Colloids in water are called hydrosols. Examples: Milk, protein