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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
FAQs on Sample Solution Paper 3: Chemistry, Class 11 - Sample Papers for Class 11 Medical and Non-Medical - JEE
| 1. What is the atomic number of an element? |  |
Ans. The atomic number of an element represents the number of protons in the nucleus of an atom of that element. It is a unique identifier for each element on the periodic table. For example, the atomic number of hydrogen is 1, which means it has one proton in its nucleus.
| 2. How is the atomic mass of an element calculated? | |
Ans. The atomic mass of an element is calculated by taking into account the mass of all the isotopes of that element and their relative abundances. It is usually expressed in atomic mass units (amu). The atomic mass can be found on the periodic table below the element's symbol.
| 3. What is the difference between a physical change and a chemical change? | |
Ans. A physical change refers to a change in the physical properties of a substance, such as its shape, size, or state of matter, without altering its chemical composition. On the other hand, a chemical change involves a transformation of substances into new substances with different chemical properties. Examples of physical changes include melting ice or boiling water, while burning wood or rusting iron are examples of chemical changes.
| 4. How is the valency of an element determined? | |
Ans. The valency of an element is determined by the number of electrons an atom of that element can gain, lose, or share to achieve a stable electron configuration. It is usually based on the number of valence electrons present in the outermost energy level of the atom. For example, elements in Group 1 of the periodic table have a valency of +1, as they tend to lose one electron to achieve a stable configuration.
| 5. What is the difference between an element and a compound? | |
Ans. An element is a pure substance that consists of only one type of atom. It cannot be broken down into simpler substances by ordinary chemical means. In contrast, a compound is a substance composed of two or more different elements chemically combined in fixed ratios. Compounds have properties that are different from the elements they are made of. For example, water (H2O) is a compound made up of two hydrogen atoms and one oxygen atom.
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