Sample Solution Paper 3 - Chemistry, Class 12 NEET Notes | EduRev

Sample Papers for Class 12 Medical and Non-Medical

Created by: Mohit Rajpoot

NEET : Sample Solution Paper 3 - Chemistry, Class 12 NEET Notes | EduRev

 Page 1


  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
CBSE 
Class XII Chemistry 
Sample Paper 3 - Solution 
 
Time: 3 Hrs                                                                     Maximum Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
 
1. A polymer formed by the polymerisation of two or more different monomers is called a 
copolymer; for example, Buna-S. 
                                          
2. This is because amorphous solids have a tendency to flow.  
     
 
                                                                                                                             
 
OR 
 
Frenkel defect is shown by ionic compounds having large difference in the size of ions 
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In 
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy 
interstitial sites.  
  
3. Pyridinium chlorochromate
                                                                                                                                 
 
 
4. Yes, if one of the reactants is taken in excess, its concentration does not change 
considerably with time. So, a bimolecular reaction can be of the first order.                                                                                  
 
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be 
used for drying ammonia.  
OR 
 
The increasing order of acidic character of haloacids is 
HF < HCl < HBr < HI 
 
Section B 
6.  
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the 
number of moles of that component to the total number of moles of all the 
components of the mixture. 
Mole fraction is given by 
nA = 
Molesof A
Total,molesof allthecomponentsof themixtur e
 
 
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as 
 
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
 
 
OR 
Page 2


  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
CBSE 
Class XII Chemistry 
Sample Paper 3 - Solution 
 
Time: 3 Hrs                                                                     Maximum Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
 
1. A polymer formed by the polymerisation of two or more different monomers is called a 
copolymer; for example, Buna-S. 
                                          
2. This is because amorphous solids have a tendency to flow.  
     
 
                                                                                                                             
 
OR 
 
Frenkel defect is shown by ionic compounds having large difference in the size of ions 
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In 
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy 
interstitial sites.  
  
3. Pyridinium chlorochromate
                                                                                                                                 
 
 
4. Yes, if one of the reactants is taken in excess, its concentration does not change 
considerably with time. So, a bimolecular reaction can be of the first order.                                                                                  
 
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be 
used for drying ammonia.  
OR 
 
The increasing order of acidic character of haloacids is 
HF < HCl < HBr < HI 
 
Section B 
6.  
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the 
number of moles of that component to the total number of moles of all the 
components of the mixture. 
Mole fraction is given by 
nA = 
Molesof A
Total,molesof allthecomponentsof themixtur e
 
 
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as 
 
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
 
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
The relationship between mole fraction and vapour pressure of a component of 
an ideal solution in the liquid phase and vapour phase: 
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
 
 
 
7.  
 
                 When Kc < 1, taking log, gives a negative value.  
                 For example: 
 
                       
 
                  Thus,   is negative if the equilibrium constant Kc < 1.                 
 
 
               
 
 
8. Given: 
Rate constant K = 200 s
-1
 
The half-life of the first-order reaction is  
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ? 
 
9.  
(a) Cl2O:   
2x – 2 = 0 
              2x = 2 
                x = + 1 
 
(b) KBrO3: 
1 + x - 6 = 0  
           x = 6 + 1  
           x = 7 
cell c
0.059
For a cell, E =  log K
n
?
cell
0.059
E =  log 0.01 
n
 = - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1     
?
?
?
?
?
Page 3


  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
CBSE 
Class XII Chemistry 
Sample Paper 3 - Solution 
 
Time: 3 Hrs                                                                     Maximum Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
 
1. A polymer formed by the polymerisation of two or more different monomers is called a 
copolymer; for example, Buna-S. 
                                          
2. This is because amorphous solids have a tendency to flow.  
     
 
                                                                                                                             
 
OR 
 
Frenkel defect is shown by ionic compounds having large difference in the size of ions 
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In 
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy 
interstitial sites.  
  
3. Pyridinium chlorochromate
                                                                                                                                 
 
 
4. Yes, if one of the reactants is taken in excess, its concentration does not change 
considerably with time. So, a bimolecular reaction can be of the first order.                                                                                  
 
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be 
used for drying ammonia.  
OR 
 
The increasing order of acidic character of haloacids is 
HF < HCl < HBr < HI 
 
Section B 
6.  
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the 
number of moles of that component to the total number of moles of all the 
components of the mixture. 
Mole fraction is given by 
nA = 
Molesof A
Total,molesof allthecomponentsof themixtur e
 
 
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as 
 
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
 
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
The relationship between mole fraction and vapour pressure of a component of 
an ideal solution in the liquid phase and vapour phase: 
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
 
 
 
7.  
 
                 When Kc < 1, taking log, gives a negative value.  
                 For example: 
 
                       
 
                  Thus,   is negative if the equilibrium constant Kc < 1.                 
 
 
               
 
 
8. Given: 
Rate constant K = 200 s
-1
 
The half-life of the first-order reaction is  
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ? 
 
9.  
(a) Cl2O:   
2x – 2 = 0 
              2x = 2 
                x = + 1 
 
(b) KBrO3: 
1 + x - 6 = 0  
           x = 6 + 1  
           x = 7 
cell c
0.059
For a cell, E =  log K
n
?
cell
0.059
E =  log 0.01 
n
 = - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1     
?
?
?
?
?
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of 
iodoform, while methanol will not.                                                                                                  
                  C2H5OH + 4I2 + 6NaOH   CHI3 + 5NaI + 5H2O + HCOONa          
                                                Yellow                                                    
                 CH3OH + I2 + NaOH  no yellow ppt. 
 
11.  
(a) Methyl-2-aminobutanoate           
                                       
(b) N-methylbenzenamine   
                                                       
OR 
 
(a) Ambident nucleophile 
(i) 
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ? 
(ii) 
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ? 
From the above reactions, it is clear that CN
-
 is an ambident nucleophile. 
 
(b) Hinsberg test: 
 
 
 
?
?
Page 4


  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
CBSE 
Class XII Chemistry 
Sample Paper 3 - Solution 
 
Time: 3 Hrs                                                                     Maximum Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
 
1. A polymer formed by the polymerisation of two or more different monomers is called a 
copolymer; for example, Buna-S. 
                                          
2. This is because amorphous solids have a tendency to flow.  
     
 
                                                                                                                             
 
OR 
 
Frenkel defect is shown by ionic compounds having large difference in the size of ions 
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In 
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy 
interstitial sites.  
  
3. Pyridinium chlorochromate
                                                                                                                                 
 
 
4. Yes, if one of the reactants is taken in excess, its concentration does not change 
considerably with time. So, a bimolecular reaction can be of the first order.                                                                                  
 
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be 
used for drying ammonia.  
OR 
 
The increasing order of acidic character of haloacids is 
HF < HCl < HBr < HI 
 
Section B 
6.  
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the 
number of moles of that component to the total number of moles of all the 
components of the mixture. 
Mole fraction is given by 
nA = 
Molesof A
Total,molesof allthecomponentsof themixtur e
 
 
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as 
 
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
 
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
The relationship between mole fraction and vapour pressure of a component of 
an ideal solution in the liquid phase and vapour phase: 
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
 
 
 
7.  
 
                 When Kc < 1, taking log, gives a negative value.  
                 For example: 
 
                       
 
                  Thus,   is negative if the equilibrium constant Kc < 1.                 
 
 
               
 
 
8. Given: 
Rate constant K = 200 s
-1
 
The half-life of the first-order reaction is  
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ? 
 
9.  
(a) Cl2O:   
2x – 2 = 0 
              2x = 2 
                x = + 1 
 
(b) KBrO3: 
1 + x - 6 = 0  
           x = 6 + 1  
           x = 7 
cell c
0.059
For a cell, E =  log K
n
?
cell
0.059
E =  log 0.01 
n
 = - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1     
?
?
?
?
?
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of 
iodoform, while methanol will not.                                                                                                  
                  C2H5OH + 4I2 + 6NaOH   CHI3 + 5NaI + 5H2O + HCOONa          
                                                Yellow                                                    
                 CH3OH + I2 + NaOH  no yellow ppt. 
 
11.  
(a) Methyl-2-aminobutanoate           
                                       
(b) N-methylbenzenamine   
                                                       
OR 
 
(a) Ambident nucleophile 
(i) 
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ? 
(ii) 
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ? 
From the above reactions, it is clear that CN
-
 is an ambident nucleophile. 
 
(b) Hinsberg test: 
 
 
 
?
?
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
12.  
(a) Benzoyl peroxide acts as an initiator because it generates free radicals.  
                                                                                                   
(b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene. 
LDPE: It is produced by free radical polymerisation at a high temperature of 350–
570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer.  
                                                                                               
HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta 
catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a 
linear polymer.                                                                           
 
Section C 
 
13. Given: 
Cell edge (a) = 288 pm = 288 ×10
-10
 cm 
Let’s first find the volume, 
Volume of unit cell = (288 ×10
-10
 cm)
3
 
                                     = 2.389×10
-23
 cm
3
 
 
3
Mass
Volume of 208 g of the element 
Density
208
7.2
28.89cm
?
?
?
 
 
   
23
23 3
Totalvolume
Numberof unit cells 
Volumeof aunit cell
28.89
2.389 10
12.09 10 cm
?
?
?
?
?
??
                        
          
For a b.c.c. structure, the number of atoms per unit cell = 2 
 
Number of atoms present in 208 g 
= No. of atoms per unit cell ×  No. of unit cells 
= 2 × 12.09 × 10
23
 
= 24.18 × 10
23
 
= 2.418×10
24
 atoms 
                         
 
 
 
 
 
 
 
Page 5


  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
CBSE 
Class XII Chemistry 
Sample Paper 3 - Solution 
 
Time: 3 Hrs                                                                     Maximum Marks: 70 
_________________________________________________________________________________________________________ 
 
Section A 
 
1. A polymer formed by the polymerisation of two or more different monomers is called a 
copolymer; for example, Buna-S. 
                                          
2. This is because amorphous solids have a tendency to flow.  
     
 
                                                                                                                             
 
OR 
 
Frenkel defect is shown by ionic compounds having large difference in the size of ions 
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In 
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy 
interstitial sites.  
  
3. Pyridinium chlorochromate
                                                                                                                                 
 
 
4. Yes, if one of the reactants is taken in excess, its concentration does not change 
considerably with time. So, a bimolecular reaction can be of the first order.                                                                                  
 
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be 
used for drying ammonia.  
OR 
 
The increasing order of acidic character of haloacids is 
HF < HCl < HBr < HI 
 
Section B 
6.  
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the 
number of moles of that component to the total number of moles of all the 
components of the mixture. 
Mole fraction is given by 
nA = 
Molesof A
Total,molesof allthecomponentsof themixtur e
 
 
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as 
 
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
 
 
OR 
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
The relationship between mole fraction and vapour pressure of a component of 
an ideal solution in the liquid phase and vapour phase: 
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
 
 
 
7.  
 
                 When Kc < 1, taking log, gives a negative value.  
                 For example: 
 
                       
 
                  Thus,   is negative if the equilibrium constant Kc < 1.                 
 
 
               
 
 
8. Given: 
Rate constant K = 200 s
-1
 
The half-life of the first-order reaction is  
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ? 
 
9.  
(a) Cl2O:   
2x – 2 = 0 
              2x = 2 
                x = + 1 
 
(b) KBrO3: 
1 + x - 6 = 0  
           x = 6 + 1  
           x = 7 
cell c
0.059
For a cell, E =  log K
n
?
cell
0.059
E =  log 0.01 
n
 = - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1     
?
?
?
?
?
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of 
iodoform, while methanol will not.                                                                                                  
                  C2H5OH + 4I2 + 6NaOH   CHI3 + 5NaI + 5H2O + HCOONa          
                                                Yellow                                                    
                 CH3OH + I2 + NaOH  no yellow ppt. 
 
11.  
(a) Methyl-2-aminobutanoate           
                                       
(b) N-methylbenzenamine   
                                                       
OR 
 
(a) Ambident nucleophile 
(i) 
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ? 
(ii) 
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ? 
From the above reactions, it is clear that CN
-
 is an ambident nucleophile. 
 
(b) Hinsberg test: 
 
 
 
?
?
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
12.  
(a) Benzoyl peroxide acts as an initiator because it generates free radicals.  
                                                                                                   
(b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene. 
LDPE: It is produced by free radical polymerisation at a high temperature of 350–
570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer.  
                                                                                               
HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta 
catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a 
linear polymer.                                                                           
 
Section C 
 
13. Given: 
Cell edge (a) = 288 pm = 288 ×10
-10
 cm 
Let’s first find the volume, 
Volume of unit cell = (288 ×10
-10
 cm)
3
 
                                     = 2.389×10
-23
 cm
3
 
 
3
Mass
Volume of 208 g of the element 
Density
208
7.2
28.89cm
?
?
?
 
 
   
23
23 3
Totalvolume
Numberof unit cells 
Volumeof aunit cell
28.89
2.389 10
12.09 10 cm
?
?
?
?
?
??
                        
          
For a b.c.c. structure, the number of atoms per unit cell = 2 
 
Number of atoms present in 208 g 
= No. of atoms per unit cell ×  No. of unit cells 
= 2 × 12.09 × 10
23
 
= 24.18 × 10
23
 
= 2.418×10
24
 atoms 
                         
 
 
 
 
 
 
 
  
 
CBSE XII  |  Chemistry  
Sample Paper  3 - Solution 
 
     
14. Given: 
1
1
1
2
1
1
1
0
1 1 2 2 1
0
1 1 2 2 1
0
11
0
1
1
0
1
1
0
1
1
0
1
1
1
W 90g
W 10g
M 72 12 96 180g mol
P 32.8mmHg
W 18g mol
P P n W M
P n n M W
PP 92 18
P 180 90
P 92 18
1
P 180 90
P 92 18
1
P 180 90
P 92 18
1
P 180 90
P 92 18
1
32.8 180 90
P 0.36KPa
?
?
?
?
? ? ? ?
?
?
?
? ? ?
?
?
??
? ? ?
?
? ? ?
?
? ??
??
??
?
??
?
??
?
?
 
 
OR 
8 18
o
oo
s
B
B
1
CH 1
80 P
P 80%of P 0.80P
100
W
n mol
40
114g
n 1mol...................(Molar mass 114gmol)
114gmol
?
?
?
? ? ?
? ? ?
 
8 18
B
os
B
B o
B
B C H
w
n pp
40
Molesof octane x
W
p n n
40 1
?
? ? ? ? ?
?
?
 
B
oo
o
B
B
B
w
p 0.80P
40
W
P
40 1
w
W
40
1 0.80
W
W 40
40 1
?
?
?
? ? ?
?
?
            
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of NEET

Dynamic Test

Content Category

Related Searches

Previous Year Questions with Solutions

,

ppt

,

Semester Notes

,

Exam

,

Extra Questions

,

Class 12 NEET Notes | EduRev

,

past year papers

,

study material

,

pdf

,

practice quizzes

,

MCQs

,

Free

,

Sample Solution Paper 3 - Chemistry

,

Objective type Questions

,

Sample Solution Paper 3 - Chemistry

,

Class 12 NEET Notes | EduRev

,

Sample Solution Paper 3 - Chemistry

,

Important questions

,

video lectures

,

shortcuts and tricks

,

mock tests for examination

,

Summary

,

Class 12 NEET Notes | EduRev

,

Viva Questions

,

Sample Paper

;