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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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FAQs on Sample Solution Paper 1 - Chemistry, Class 12 - Additional Study Material for NEET
| 1. What is the structure of an atom? |  |
Ans. An atom is composed of a central nucleus made up of protons and neutrons, surrounded by electrons in energy levels or shells. The protons and neutrons are located in the nucleus, whereas the electrons orbit around the nucleus in specific energy levels.
| 2. 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, while a compound is a substance that contains two or more different types of elements chemically bonded together. Elements cannot be broken down into simpler substances, whereas compounds can be broken down into their constituent elements.
| 3. How are chemical reactions represented? | |
Ans. Chemical reactions are represented using chemical equations, with reactants on the left side and products on the right side. The reactants are the substances that undergo a chemical change, and the products are the substances formed as a result of the reaction. The chemical equation also includes coefficients, which represent the relative amounts of each substance involved in the reaction.
| 4. What is the difference between an exothermic and an endothermic reaction? | |
Ans. An exothermic reaction is a chemical reaction that releases heat energy to the surroundings. In contrast, an endothermic reaction absorbs heat energy from the surroundings. Exothermic reactions often feel warm to the touch, while endothermic reactions feel cool.
| 5. How does the periodic table organize elements? | |
Ans. The periodic table organizes elements based on their atomic number, which is the number of protons in the nucleus of an atom. Elements are arranged in order of increasing atomic number from left to right and top to bottom. The periodic table also groups elements with similar properties into columns called groups or families.
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Sample Solution Paper 1 - Chemistry, Class 12 Notes
Sample Solution Paper 1 - Chemistry, Class 12 Notes offer in-depth insights into the specific topic to help you master it with ease.
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Sample Solution Paper 1 - Chemistry, Class 12 NEET Questions
The "Sample Solution Paper 1 - Chemistry, Class 12 NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
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The content of Sample Solution Paper 1 - Chemistry, Class 12 is prepared as per the latest NEET syllabus.
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