Sample Solution Paper 3 - Chemistry, Class 11 NEET Notes | EduRev

Sample Papers for Class 11 Medical and Non-Medical

NEET : Sample Solution Paper 3 - Chemistry, Class 11 NEET Notes | EduRev

 Page 1


  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
CBSE 
Class XI Chemistry 
Sample Paper – 3 Solution 
 
  Section A 
1. HI>HBr>HCl>HF 
 
2. BeCl2: Linear          
SiCl4: Tetrahedral 
OR 
 
The electron pairs involved in the bond formation are known as bond pairs or shared 
pairs. 
 
3. IUPAC name of allyl alcohol: Prop-2-en-1-ol 
 
4.  The amount of oxygen required by bacteria to breakdown the organic matter present in 
a certain volume of a sample of water is called biochemical oxygen demand. 
    OR 
Carboxyhaemolobin is the compound formed when CO combines with blood. 
 
5. Sodium (Na) – Yellow 
Poassium (K) - Violet 
 
  Section B 
  
6. Metallic character decreases and non metallic character increases in moving from left to 
right in a period. This is due to increase in ionization enthalpy and electron gain 
enthalpy. 
 
7. Increasing order of size: Al
3+
    <    Mg
2+
    <    Na
+
   <    F
- 
  <   O
2-
                      
This is an isoelectronic series i.e. the number electrons are the same in all the elements. 
Thus, as the effective nuclear charge decreases, electrons are held away from the 
nucleus and thus size increases.   
     
8. Given: 
Velocity of electron = 2.07×10
7
 m/s 
Mass of electron = 9.1×10
-31
 kg 
We know, 
Page 2


  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
CBSE 
Class XI Chemistry 
Sample Paper – 3 Solution 
 
  Section A 
1. HI>HBr>HCl>HF 
 
2. BeCl2: Linear          
SiCl4: Tetrahedral 
OR 
 
The electron pairs involved in the bond formation are known as bond pairs or shared 
pairs. 
 
3. IUPAC name of allyl alcohol: Prop-2-en-1-ol 
 
4.  The amount of oxygen required by bacteria to breakdown the organic matter present in 
a certain volume of a sample of water is called biochemical oxygen demand. 
    OR 
Carboxyhaemolobin is the compound formed when CO combines with blood. 
 
5. Sodium (Na) – Yellow 
Poassium (K) - Violet 
 
  Section B 
  
6. Metallic character decreases and non metallic character increases in moving from left to 
right in a period. This is due to increase in ionization enthalpy and electron gain 
enthalpy. 
 
7. Increasing order of size: Al
3+
    <    Mg
2+
    <    Na
+
   <    F
- 
  <   O
2-
                      
This is an isoelectronic series i.e. the number electrons are the same in all the elements. 
Thus, as the effective nuclear charge decreases, electrons are held away from the 
nucleus and thus size increases.   
     
8. Given: 
Velocity of electron = 2.07×10
7
 m/s 
Mass of electron = 9.1×10
-31
 kg 
We know, 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
34
31 7
11
h
mv
6.63 10
9.1 10 2.05 0.5 10
3.55 10 m
?
?
?
??
?
?
? ? ? ?
??
 
9. Given: 
Pressure P = 5 bar 
Molar mass of nitrogen M = 28 g/mol 
Density of nitrogen, 
? ?
PM
RT
Density of gaseousoxideis,
0.987 5 x
0.987 5 28
273 0.0821 273 0.0821
0.987 5 28 273 0.0821
X
273 0.0821 0..987 5
70
??
??
??
?
??
? ? ? ?
?
? ? ?
?
 
The molar mass of the oxide is 70 g/mol 
10. 2H2O+ 2 F2 ?  4HF+ O2 
In this reaction H2O is getting oxidized to O2 and F2 is getting reduced to F
-
 ion. 
Therefore, F2 is the oxidizing agent and H2O is reducing agent.  
OR 
(a) PbS(g) + H2O2(aq)   ? PbSO 4(s) + 4H2O(l) 
 
(b) CO(g) + 2 H2(g) 
Cobalt
catalyst
? ? ? ? CH 3OH 
 
11. Given: 
No. Nuclei No. of protons No. of neutrons 
1 
56
26
Fe 26 30 
2 
88
38
Sr 38 50 
 
 
 
Page 3


  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
CBSE 
Class XI Chemistry 
Sample Paper – 3 Solution 
 
  Section A 
1. HI>HBr>HCl>HF 
 
2. BeCl2: Linear          
SiCl4: Tetrahedral 
OR 
 
The electron pairs involved in the bond formation are known as bond pairs or shared 
pairs. 
 
3. IUPAC name of allyl alcohol: Prop-2-en-1-ol 
 
4.  The amount of oxygen required by bacteria to breakdown the organic matter present in 
a certain volume of a sample of water is called biochemical oxygen demand. 
    OR 
Carboxyhaemolobin is the compound formed when CO combines with blood. 
 
5. Sodium (Na) – Yellow 
Poassium (K) - Violet 
 
  Section B 
  
6. Metallic character decreases and non metallic character increases in moving from left to 
right in a period. This is due to increase in ionization enthalpy and electron gain 
enthalpy. 
 
7. Increasing order of size: Al
3+
    <    Mg
2+
    <    Na
+
   <    F
- 
  <   O
2-
                      
This is an isoelectronic series i.e. the number electrons are the same in all the elements. 
Thus, as the effective nuclear charge decreases, electrons are held away from the 
nucleus and thus size increases.   
     
8. Given: 
Velocity of electron = 2.07×10
7
 m/s 
Mass of electron = 9.1×10
-31
 kg 
We know, 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
34
31 7
11
h
mv
6.63 10
9.1 10 2.05 0.5 10
3.55 10 m
?
?
?
??
?
?
? ? ? ?
??
 
9. Given: 
Pressure P = 5 bar 
Molar mass of nitrogen M = 28 g/mol 
Density of nitrogen, 
? ?
PM
RT
Density of gaseousoxideis,
0.987 5 x
0.987 5 28
273 0.0821 273 0.0821
0.987 5 28 273 0.0821
X
273 0.0821 0..987 5
70
??
??
??
?
??
? ? ? ?
?
? ? ?
?
 
The molar mass of the oxide is 70 g/mol 
10. 2H2O+ 2 F2 ?  4HF+ O2 
In this reaction H2O is getting oxidized to O2 and F2 is getting reduced to F
-
 ion. 
Therefore, F2 is the oxidizing agent and H2O is reducing agent.  
OR 
(a) PbS(g) + H2O2(aq)   ? PbSO 4(s) + 4H2O(l) 
 
(b) CO(g) + 2 H2(g) 
Cobalt
catalyst
? ? ? ? CH 3OH 
 
11. Given: 
No. Nuclei No. of protons No. of neutrons 
1 
56
26
Fe 26 30 
2 
88
38
Sr 38 50 
 
 
 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
12. Molar mass of methanol (CH3OH) = 32 g/mol 
                                                                  = 0.032 kg/mol 
Molarity of solution = 
 
0.793
0.032
24.78mol / l
?
?
 
We have, 
M1V1 = M2V2 
24.78 × V1 = 0.25 × 2.5 
V1 = 25.22 ml 
The required volume is 25.22 ml 
OR 
Given: 
M= 3 mol/lit 
Mass of NaCl in 1 litre solution = 3×58.5 
                                                       = 175.5 g 
Mass of water in solution = 1000 × 1.25  
                                                  = 1250 g 
Mass of water in solution = 1250- 175.5  
                                                  = 1074.5 g 
 
Molarity  
Number of moles of solute
Massof solvent in kg
3
1.074
2.79m
?
?
?
 
 
 
 
 
 
 
  
Page 4


  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
CBSE 
Class XI Chemistry 
Sample Paper – 3 Solution 
 
  Section A 
1. HI>HBr>HCl>HF 
 
2. BeCl2: Linear          
SiCl4: Tetrahedral 
OR 
 
The electron pairs involved in the bond formation are known as bond pairs or shared 
pairs. 
 
3. IUPAC name of allyl alcohol: Prop-2-en-1-ol 
 
4.  The amount of oxygen required by bacteria to breakdown the organic matter present in 
a certain volume of a sample of water is called biochemical oxygen demand. 
    OR 
Carboxyhaemolobin is the compound formed when CO combines with blood. 
 
5. Sodium (Na) – Yellow 
Poassium (K) - Violet 
 
  Section B 
  
6. Metallic character decreases and non metallic character increases in moving from left to 
right in a period. This is due to increase in ionization enthalpy and electron gain 
enthalpy. 
 
7. Increasing order of size: Al
3+
    <    Mg
2+
    <    Na
+
   <    F
- 
  <   O
2-
                      
This is an isoelectronic series i.e. the number electrons are the same in all the elements. 
Thus, as the effective nuclear charge decreases, electrons are held away from the 
nucleus and thus size increases.   
     
8. Given: 
Velocity of electron = 2.07×10
7
 m/s 
Mass of electron = 9.1×10
-31
 kg 
We know, 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
34
31 7
11
h
mv
6.63 10
9.1 10 2.05 0.5 10
3.55 10 m
?
?
?
??
?
?
? ? ? ?
??
 
9. Given: 
Pressure P = 5 bar 
Molar mass of nitrogen M = 28 g/mol 
Density of nitrogen, 
? ?
PM
RT
Density of gaseousoxideis,
0.987 5 x
0.987 5 28
273 0.0821 273 0.0821
0.987 5 28 273 0.0821
X
273 0.0821 0..987 5
70
??
??
??
?
??
? ? ? ?
?
? ? ?
?
 
The molar mass of the oxide is 70 g/mol 
10. 2H2O+ 2 F2 ?  4HF+ O2 
In this reaction H2O is getting oxidized to O2 and F2 is getting reduced to F
-
 ion. 
Therefore, F2 is the oxidizing agent and H2O is reducing agent.  
OR 
(a) PbS(g) + H2O2(aq)   ? PbSO 4(s) + 4H2O(l) 
 
(b) CO(g) + 2 H2(g) 
Cobalt
catalyst
? ? ? ? CH 3OH 
 
11. Given: 
No. Nuclei No. of protons No. of neutrons 
1 
56
26
Fe 26 30 
2 
88
38
Sr 38 50 
 
 
 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
12. Molar mass of methanol (CH3OH) = 32 g/mol 
                                                                  = 0.032 kg/mol 
Molarity of solution = 
 
0.793
0.032
24.78mol / l
?
?
 
We have, 
M1V1 = M2V2 
24.78 × V1 = 0.25 × 2.5 
V1 = 25.22 ml 
The required volume is 25.22 ml 
OR 
Given: 
M= 3 mol/lit 
Mass of NaCl in 1 litre solution = 3×58.5 
                                                       = 175.5 g 
Mass of water in solution = 1000 × 1.25  
                                                  = 1250 g 
Mass of water in solution = 1250- 175.5  
                                                  = 1074.5 g 
 
Molarity  
Number of moles of solute
Massof solvent in kg
3
1.074
2.79m
?
?
?
 
 
 
 
 
 
 
  
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
Section C 
 
13.  
      We know, 
      
? ?
? ?
? ?
? ?
18 2
n 2
18 2
1 2
18
2
n
2
n
2.18 10 Z
E
n
For He ,
n 1, n 2
2.18 10 2
E
1
8.72 10 J
0.0529 n
r
Z
Here,n 1, Z 2
0.0529 1
r
2
0.02645nm
?
?
?
?
??
?
??
??
?
? ? ?
?
??
?
?
 
Energy is 8.72 ×10
-18
 J 
Radius of the orbit is 0.02645 nm 
OR 
Let us find out the atomic number of element. 
Let the number of protons = x 
? ?
x 31.7
Number of neutrons x
100
x 0.317x
Massno.of element No.of protons No.of neutrons
81 x x 0.317x
81 2.317x
x 35
No.of protons 35
No. of neutrons 81 35
46
Atomic numer of element is 35
?
??
??
??
? ? ?
?
?
??
??
?
 
The element with atomic number 35 is bromine (Br). 
Page 5


  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
CBSE 
Class XI Chemistry 
Sample Paper – 3 Solution 
 
  Section A 
1. HI>HBr>HCl>HF 
 
2. BeCl2: Linear          
SiCl4: Tetrahedral 
OR 
 
The electron pairs involved in the bond formation are known as bond pairs or shared 
pairs. 
 
3. IUPAC name of allyl alcohol: Prop-2-en-1-ol 
 
4.  The amount of oxygen required by bacteria to breakdown the organic matter present in 
a certain volume of a sample of water is called biochemical oxygen demand. 
    OR 
Carboxyhaemolobin is the compound formed when CO combines with blood. 
 
5. Sodium (Na) – Yellow 
Poassium (K) - Violet 
 
  Section B 
  
6. Metallic character decreases and non metallic character increases in moving from left to 
right in a period. This is due to increase in ionization enthalpy and electron gain 
enthalpy. 
 
7. Increasing order of size: Al
3+
    <    Mg
2+
    <    Na
+
   <    F
- 
  <   O
2-
                      
This is an isoelectronic series i.e. the number electrons are the same in all the elements. 
Thus, as the effective nuclear charge decreases, electrons are held away from the 
nucleus and thus size increases.   
     
8. Given: 
Velocity of electron = 2.07×10
7
 m/s 
Mass of electron = 9.1×10
-31
 kg 
We know, 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
34
31 7
11
h
mv
6.63 10
9.1 10 2.05 0.5 10
3.55 10 m
?
?
?
??
?
?
? ? ? ?
??
 
9. Given: 
Pressure P = 5 bar 
Molar mass of nitrogen M = 28 g/mol 
Density of nitrogen, 
? ?
PM
RT
Density of gaseousoxideis,
0.987 5 x
0.987 5 28
273 0.0821 273 0.0821
0.987 5 28 273 0.0821
X
273 0.0821 0..987 5
70
??
??
??
?
??
? ? ? ?
?
? ? ?
?
 
The molar mass of the oxide is 70 g/mol 
10. 2H2O+ 2 F2 ?  4HF+ O2 
In this reaction H2O is getting oxidized to O2 and F2 is getting reduced to F
-
 ion. 
Therefore, F2 is the oxidizing agent and H2O is reducing agent.  
OR 
(a) PbS(g) + H2O2(aq)   ? PbSO 4(s) + 4H2O(l) 
 
(b) CO(g) + 2 H2(g) 
Cobalt
catalyst
? ? ? ? CH 3OH 
 
11. Given: 
No. Nuclei No. of protons No. of neutrons 
1 
56
26
Fe 26 30 
2 
88
38
Sr 38 50 
 
 
 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
12. Molar mass of methanol (CH3OH) = 32 g/mol 
                                                                  = 0.032 kg/mol 
Molarity of solution = 
 
0.793
0.032
24.78mol / l
?
?
 
We have, 
M1V1 = M2V2 
24.78 × V1 = 0.25 × 2.5 
V1 = 25.22 ml 
The required volume is 25.22 ml 
OR 
Given: 
M= 3 mol/lit 
Mass of NaCl in 1 litre solution = 3×58.5 
                                                       = 175.5 g 
Mass of water in solution = 1000 × 1.25  
                                                  = 1250 g 
Mass of water in solution = 1250- 175.5  
                                                  = 1074.5 g 
 
Molarity  
Number of moles of solute
Massof solvent in kg
3
1.074
2.79m
?
?
?
 
 
 
 
 
 
 
  
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
Section C 
 
13.  
      We know, 
      
? ?
? ?
? ?
? ?
18 2
n 2
18 2
1 2
18
2
n
2
n
2.18 10 Z
E
n
For He ,
n 1, n 2
2.18 10 2
E
1
8.72 10 J
0.0529 n
r
Z
Here,n 1, Z 2
0.0529 1
r
2
0.02645nm
?
?
?
?
??
?
??
??
?
? ? ?
?
??
?
?
 
Energy is 8.72 ×10
-18
 J 
Radius of the orbit is 0.02645 nm 
OR 
Let us find out the atomic number of element. 
Let the number of protons = x 
? ?
x 31.7
Number of neutrons x
100
x 0.317x
Massno.of element No.of protons No.of neutrons
81 x x 0.317x
81 2.317x
x 35
No.of protons 35
No. of neutrons 81 35
46
Atomic numer of element is 35
?
??
??
??
? ? ?
?
?
??
??
?
 
The element with atomic number 35 is bromine (Br). 
  
 
CBSE XI | Chemistry 
Sample Paper – 3 Solution 
 
    
14. The balanced chemical equation is  
              
22
2CO O 2CO
2mol 1mol
2x22.4L 22.4L
? ? ? ?
                               
Volume of oxygen required to convert 2 x 22.4 L of CO at N.T.P. = 22.4 L 
Volume of oxygen required to convert 5.2 L of CO at N.T.P. = xL
x
22.4
5.2 2.6
2 22.4
? 
OR 
 
(i) H2
+
 is more stable than H2
- 
as it contains no electron in antibonding MO while latter 
contains an electron in antibonding MO making it less st 
(ii)  PCl5 contains axial and equatorial bonds. Axial bonds are longer than equatorial 
bonds as they face more repulsion from equatorial bonds. Hence axial bonds are 
weaker than equatorial bonds.                                                                                                                             
(iii)  NaI is more covalent due to high polarizability of iodide ion due to its bigger size 
than chloride ion.                                                                                    
15.  
(i) This is due to the reason that the molecules which undergo evaporation are high 
energy molecules and therefore, the kinetic energy of the remaining molecules 
becomes less. Since the remaining molecules have lower average kinetic energy, 
their temperature becomes low.                                                                             
(ii) This is due to surface tension of liquids. Due to surface tension, the molecules of a 
liquid, try to make surface area to be minimum and for a given volume, sphere has 
the minimum surface area. Therefore the falling liquid drops are spherical.  
(iii) Intermolecular forces are stronger in acetone than in ether. Thus the vapour 
pressure of acetone is less than ether.    
  
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Extra Questions

,

Previous Year Questions with Solutions

,

Class 11 NEET Notes | EduRev

,

Sample Solution Paper 3 - Chemistry

,

Sample Solution Paper 3 - Chemistry

,

Objective type Questions

,

past year papers

,

Summary

,

Semester Notes

,

Class 11 NEET Notes | EduRev

,

pdf

,

mock tests for examination

,

Exam

,

Class 11 NEET Notes | EduRev

,

Sample Paper

,

study material

,

Sample Solution Paper 3 - Chemistry

,

shortcuts and tricks

,

practice quizzes

,

MCQs

,

Viva Questions

,

ppt

,

Important questions

,

Free

,

video lectures

;