Grade 12 Exam > Grade 12 Notes > Chemistry for Grade 12 > Chemistry: Sample Solution Paper- 4, Class 12
Chemistry: Sample Solution Paper- 4, Class 12 | Chemistry for Grade 12 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
CBSE XII | Chemistry
Sample Paper 4 - Solution
CBSE
Class XII Chemistry
Sample Paper 4 - Solution
Time: 3 Hrs Total Marks: 70
_________________________________________________________________________________________________________
Section A
1. Interstitial defect
2. Brownian movement
OR
The scattering of light by colloidal particles is known as the Tyndall effect.
3. The maximum oxidation state shown by actinoids is +7.
OR
The transition element which does not exhibit variation in the oxidation state in its
compounds is zinc (Zn).
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing
or –I effect of three groups. Thus, it gives a salt with Na2CO3 which is a weak base.
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3.
5. Elastomers are polymers in which polymer chains are held together by weak
intermolecular forces and hence are stretchable. Fibres are polymers with strong
intermolecular forces like hydrogen bonding and hence possess high tensile strength.
Section B
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that the mole fraction of a gas in the
solution is proportional to the partial pressure of the gas over the solution.
p = KH . x
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these
beverages, carbon dioxide is passed at high pressure to increase its solubility.
OR
Using the relation,
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ?
Solving for m, we get m=
b.p.(solution) b.p.(l)
b.p.
TT
k
?
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
-NO
2
Page 2
CBSE XII | Chemistry
Sample Paper 4 - Solution
CBSE
Class XII Chemistry
Sample Paper 4 - Solution
Time: 3 Hrs Total Marks: 70
_________________________________________________________________________________________________________
Section A
1. Interstitial defect
2. Brownian movement
OR
The scattering of light by colloidal particles is known as the Tyndall effect.
3. The maximum oxidation state shown by actinoids is +7.
OR
The transition element which does not exhibit variation in the oxidation state in its
compounds is zinc (Zn).
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing
or –I effect of three groups. Thus, it gives a salt with Na2CO3 which is a weak base.
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3.
5. Elastomers are polymers in which polymer chains are held together by weak
intermolecular forces and hence are stretchable. Fibres are polymers with strong
intermolecular forces like hydrogen bonding and hence possess high tensile strength.
Section B
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that the mole fraction of a gas in the
solution is proportional to the partial pressure of the gas over the solution.
p = KH . x
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these
beverages, carbon dioxide is passed at high pressure to increase its solubility.
OR
Using the relation,
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ?
Solving for m, we get m=
b.p.(solution) b.p.(l)
b.p.
TT
k
?
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
-NO
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
7. Fe + 2H
+
? Fe
2+
+ H2
8. Hexagonal
Example: Graphite
9.
(a)
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ?
(b)
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ?
10. Compound (a) reacts faster than compound (b).
This is due to the formation of a more stable carbocation in compound (a) in the
rate-determining step than carbocation in compound (b).
11. The increasing order of the acidic character of the given species is as follows:
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH
OR
(a) pent-2-enal
(b) 3-phenyl prop-2-enol
12.
(a) Monomers of Dacron:
ethylene glycol and terephthalic acid
(b) Monomers of Buna-N:
1, 3-butadiene and acrylonitrile
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
Page 3
CBSE XII | Chemistry
Sample Paper 4 - Solution
CBSE
Class XII Chemistry
Sample Paper 4 - Solution
Time: 3 Hrs Total Marks: 70
_________________________________________________________________________________________________________
Section A
1. Interstitial defect
2. Brownian movement
OR
The scattering of light by colloidal particles is known as the Tyndall effect.
3. The maximum oxidation state shown by actinoids is +7.
OR
The transition element which does not exhibit variation in the oxidation state in its
compounds is zinc (Zn).
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing
or –I effect of three groups. Thus, it gives a salt with Na2CO3 which is a weak base.
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3.
5. Elastomers are polymers in which polymer chains are held together by weak
intermolecular forces and hence are stretchable. Fibres are polymers with strong
intermolecular forces like hydrogen bonding and hence possess high tensile strength.
Section B
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that the mole fraction of a gas in the
solution is proportional to the partial pressure of the gas over the solution.
p = KH . x
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these
beverages, carbon dioxide is passed at high pressure to increase its solubility.
OR
Using the relation,
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ?
Solving for m, we get m=
b.p.(solution) b.p.(l)
b.p.
TT
k
?
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
-NO
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
7. Fe + 2H
+
? Fe
2+
+ H2
8. Hexagonal
Example: Graphite
9.
(a)
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ?
(b)
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ?
10. Compound (a) reacts faster than compound (b).
This is due to the formation of a more stable carbocation in compound (a) in the
rate-determining step than carbocation in compound (b).
11. The increasing order of the acidic character of the given species is as follows:
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH
OR
(a) pent-2-enal
(b) 3-phenyl prop-2-enol
12.
(a) Monomers of Dacron:
ethylene glycol and terephthalic acid
(b) Monomers of Buna-N:
1, 3-butadiene and acrylonitrile
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
Section C
13. NA = 6.023 × 10
23
Atomic mass of mercury = 200
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
So, the number of atoms of Hg present in 1 g of Hg =
23
21
6.023 10
3.0115 10
200
?
??
The given density of mercury is 13.6 g/cc.
? Volume of 1 atom of mercury (Hg) =
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
Each mercury atom occupies a cube of edge length equal to its diameter.
So,
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
°
14.
T = 300 K
c = 0.1 M
i
OR
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate
of evaporation becomes equal to the rate of condensation. Equilibrium constant
does not change at a particular temperature, and therefore, the vapour pressure
remains constant.
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is
positive. Thus, the solution shows positive deviation.
(c) B will show greater lowering of vapour pressure because
0s
21
0
12
wM PP
P w M
?
?
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
Page 4
CBSE XII | Chemistry
Sample Paper 4 - Solution
CBSE
Class XII Chemistry
Sample Paper 4 - Solution
Time: 3 Hrs Total Marks: 70
_________________________________________________________________________________________________________
Section A
1. Interstitial defect
2. Brownian movement
OR
The scattering of light by colloidal particles is known as the Tyndall effect.
3. The maximum oxidation state shown by actinoids is +7.
OR
The transition element which does not exhibit variation in the oxidation state in its
compounds is zinc (Zn).
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing
or –I effect of three groups. Thus, it gives a salt with Na2CO3 which is a weak base.
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3.
5. Elastomers are polymers in which polymer chains are held together by weak
intermolecular forces and hence are stretchable. Fibres are polymers with strong
intermolecular forces like hydrogen bonding and hence possess high tensile strength.
Section B
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that the mole fraction of a gas in the
solution is proportional to the partial pressure of the gas over the solution.
p = KH . x
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these
beverages, carbon dioxide is passed at high pressure to increase its solubility.
OR
Using the relation,
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ?
Solving for m, we get m=
b.p.(solution) b.p.(l)
b.p.
TT
k
?
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
-NO
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
7. Fe + 2H
+
? Fe
2+
+ H2
8. Hexagonal
Example: Graphite
9.
(a)
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ?
(b)
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ?
10. Compound (a) reacts faster than compound (b).
This is due to the formation of a more stable carbocation in compound (a) in the
rate-determining step than carbocation in compound (b).
11. The increasing order of the acidic character of the given species is as follows:
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH
OR
(a) pent-2-enal
(b) 3-phenyl prop-2-enol
12.
(a) Monomers of Dacron:
ethylene glycol and terephthalic acid
(b) Monomers of Buna-N:
1, 3-butadiene and acrylonitrile
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
Section C
13. NA = 6.023 × 10
23
Atomic mass of mercury = 200
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
So, the number of atoms of Hg present in 1 g of Hg =
23
21
6.023 10
3.0115 10
200
?
??
The given density of mercury is 13.6 g/cc.
? Volume of 1 atom of mercury (Hg) =
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
Each mercury atom occupies a cube of edge length equal to its diameter.
So,
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
°
14.
T = 300 K
c = 0.1 M
i
OR
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate
of evaporation becomes equal to the rate of condensation. Equilibrium constant
does not change at a particular temperature, and therefore, the vapour pressure
remains constant.
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is
positive. Thus, the solution shows positive deviation.
(c) B will show greater lowering of vapour pressure because
0s
21
0
12
wM PP
P w M
?
?
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
CBSE XII | Chemistry
Sample Paper 4 - Solution
15.
(a) Observing the graph, there is a decrease in the concentration of A in the time period
of four hours from 1 hr to 5 hr, i.e. 0.5 – 0.3 = 0.2 M.
In this same period of time, an increase in the concentration of B is twice the
decrease in concentration of A. Thus, n = 2.
(b)
2
2
eq
eq
[B ]
(0.6)
k 1.2
[A ] (0.3)
? ? ?
(c) Initial rate of conversion of A = change in conc. of A during 1 hour
11
0.6 0.5
0.1mollitre hour
1
??
?
??
16.
(a) It causes peptisation leading to the formation of a reddish brown colloidal solution
of Fe(OH)3.
(b) Scattering of light by colloidal particles takes place and the path of light becomes
illuminated. This is called Tyndall effect.
(c) Under the influence of current, the colloidal particles start moving towards the
oppositely charged electrode. This is called electrophoresis.
17.
(a) Reduction reaction is feasible only at a high temperature; thus, it is not economically
and practically viable.
(b) This is done to remove basic impurities by the formation of slag.
Example:
(c) In electrometallurgy of Al, graphite rods act as the anode and get burnt away as CO
and CO2 during electrolysis.
18.
(a) Mischmetal
(b) Composition: 95% lanthanoid metal, 5% Fe and traces of S, C, Ca and Al.
(c) is more basic than .
Due to lanthanoid contraction, La is bigger in size than Lu. Larger the M-OH bond,
weaker is the M-OH bond. This makes La(OH)3 more basic.
This is due to lanthanoid contraction because of the poor screening effect of f
electrons. As a consequence of this, elements of the second and third transition
series have similar size. Therefore, Zr and Hf have similar atomic radii.
OR
?
23
FeO + SiO FeSiO
(Slag)
? ? La OH
3
? ? Lu OH
3
Page 5
CBSE XII | Chemistry
Sample Paper 4 - Solution
CBSE
Class XII Chemistry
Sample Paper 4 - Solution
Time: 3 Hrs Total Marks: 70
_________________________________________________________________________________________________________
Section A
1. Interstitial defect
2. Brownian movement
OR
The scattering of light by colloidal particles is known as the Tyndall effect.
3. The maximum oxidation state shown by actinoids is +7.
OR
The transition element which does not exhibit variation in the oxidation state in its
compounds is zinc (Zn).
4. 2, 4, 6-Trinitrophenol is more acidic than phenol because of the electron-withdrawing
or –I effect of three groups. Thus, it gives a salt with Na2CO3 which is a weak base.
So, this makes 2, 4, 6-trinitrophenol soluble in aqueous Na2CO3.
5. Elastomers are polymers in which polymer chains are held together by weak
intermolecular forces and hence are stretchable. Fibres are polymers with strong
intermolecular forces like hydrogen bonding and hence possess high tensile strength.
Section B
6. Henry’s law states that solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure of a gas. If we use mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that the mole fraction of a gas in the
solution is proportional to the partial pressure of the gas over the solution.
p = KH . x
Application: Soft drinks contain dissolved carbon dioxide. In the preparation of these
beverages, carbon dioxide is passed at high pressure to increase its solubility.
OR
Using the relation,
b.p.(solution) b.p.(l) b.p.
T T k m ? ? ?
Solving for m, we get m=
b.p.(solution) b.p.(l)
b.p.
TT
k
?
b.p. 1
-1
b.p.
?T
1.00K
m= = 1.95molkg
k 0.512Kkgmol
?
?
-NO
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
7. Fe + 2H
+
? Fe
2+
+ H2
8. Hexagonal
Example: Graphite
9.
(a)
2 2 2 2
4HCl MnO MnCl Cl 2H O ? ? ? ? ? ?
(b)
Diffusedsunlight
22
H Cl 2HCl ? ? ? ? ? ? ?
10. Compound (a) reacts faster than compound (b).
This is due to the formation of a more stable carbocation in compound (a) in the
rate-determining step than carbocation in compound (b).
11. The increasing order of the acidic character of the given species is as follows:
CF3COOH > CCl3COOH > CH2ClCOOH > HCOOH
OR
(a) pent-2-enal
(b) 3-phenyl prop-2-enol
12.
(a) Monomers of Dacron:
ethylene glycol and terephthalic acid
(b) Monomers of Buna-N:
1, 3-butadiene and acrylonitrile
?
2+
+2
0.059 [Fe ]
E = E - log
n [H ]
???
E = E - E
cathode anode
= 0 -(-0.44V)
= 0.44 V
?
0.059 0.001
E = E - log
21
0.059 0.001
= 0.44 - log
21
= 0.5285 V
? ?
o
3
? ?
o
2
? ?
HOCH CH OH
22
? ?
CH = CH - CH= CH
22
? ?
CH = CH- CN
2
CBSE XII | Chemistry
Sample Paper 4 - Solution
Section C
13. NA = 6.023 × 10
23
Atomic mass of mercury = 200
?Number of atoms present in 200 g of Hg = 6.023 × 10
23
So, the number of atoms of Hg present in 1 g of Hg =
23
21
6.023 10
3.0115 10
200
?
??
The given density of mercury is 13.6 g/cc.
? Volume of 1 atom of mercury (Hg) =
23
21
1
2.44 10 cc
3.0115 10 13.6
?
??
??
Each mercury atom occupies a cube of edge length equal to its diameter.
So,
Diameter of an atom of mercury = (2.44 ×10
-23
)
1/3
cm = 2.905 ×10
-8
cm = 2.91A
°
14.
T = 300 K
c = 0.1 M
i
OR
(a) Vapour pressure is the pressure of the vapour at the equilibrium state when the rate
of evaporation becomes equal to the rate of condensation. Equilibrium constant
does not change at a particular temperature, and therefore, the vapour pressure
remains constant.
(b) As the solution becomes cool, heat gets absorbed; hence, enthalpy change is
positive. Thus, the solution shows positive deviation.
(c) B will show greater lowering of vapour pressure because
0s
21
0
12
wM PP
P w M
?
?
p = 4 . 6 a t m n = 2
= 1.87
? ? ? ?
+-
NaCl Na + Cl
Initial 1 0 0
After
dissociation 1 - a a a
1- a + a + a
i=
1
1+ a
= = 1 + a
1
1.87 = 1 + a
a = 1 .8 7 - 1
= 0.87
CBSE XII | Chemistry
Sample Paper 4 - Solution
15.
(a) Observing the graph, there is a decrease in the concentration of A in the time period
of four hours from 1 hr to 5 hr, i.e. 0.5 – 0.3 = 0.2 M.
In this same period of time, an increase in the concentration of B is twice the
decrease in concentration of A. Thus, n = 2.
(b)
2
2
eq
eq
[B ]
(0.6)
k 1.2
[A ] (0.3)
? ? ?
(c) Initial rate of conversion of A = change in conc. of A during 1 hour
11
0.6 0.5
0.1mollitre hour
1
??
?
??
16.
(a) It causes peptisation leading to the formation of a reddish brown colloidal solution
of Fe(OH)3.
(b) Scattering of light by colloidal particles takes place and the path of light becomes
illuminated. This is called Tyndall effect.
(c) Under the influence of current, the colloidal particles start moving towards the
oppositely charged electrode. This is called electrophoresis.
17.
(a) Reduction reaction is feasible only at a high temperature; thus, it is not economically
and practically viable.
(b) This is done to remove basic impurities by the formation of slag.
Example:
(c) In electrometallurgy of Al, graphite rods act as the anode and get burnt away as CO
and CO2 during electrolysis.
18.
(a) Mischmetal
(b) Composition: 95% lanthanoid metal, 5% Fe and traces of S, C, Ca and Al.
(c) is more basic than .
Due to lanthanoid contraction, La is bigger in size than Lu. Larger the M-OH bond,
weaker is the M-OH bond. This makes La(OH)3 more basic.
This is due to lanthanoid contraction because of the poor screening effect of f
electrons. As a consequence of this, elements of the second and third transition
series have similar size. Therefore, Zr and Hf have similar atomic radii.
OR
?
23
FeO + SiO FeSiO
(Slag)
? ? La OH
3
? ? Lu OH
3
CBSE XII | Chemistry
Sample Paper 4 - Solution
(a) Cr
2+
is strongly reducing because it changes to Cr
3+
which is more stable as Cr
3+
has
higher hydration energy. Manganese (III) is strongly oxidising because it changes
from Mn(III) to Mn(II) which is more stable due to half-filled d-orbitals.
(b) Co(III) has greater tendency to form coordination complexes than Co (II). Thus, in
the presence of ligands, Co (II) changes to Co (III), i.e. it is easily oxidised.
(c) The ions in d
1
configuration have great tendency to lose one electron and acquire
d
0
configuration and hence act as reducing agents. All elements with d1
configuration are either reducing or undergo disproportionation.
Example:
10
6 7 4
2
4 4 2 2
3d 3d
3MnO 4H 2MnO MnO H O
? ? ?
? ? ?
? ? ? ? ? ?
19.
(a) Ambidentate ligand: A unidentate ligand which can bind to the central metal atom
through any of the two donor atoms present in it is called ambidentate ligand.
Example:
2
NO ,SCN
??
M SCN M NCS
Thiocynato Isothiocynato
??
(b)
(i) Potassium trioxalatochromate (III)
(ii) Diamminedichloridoplatinum (II)
(c)
20.
(a)
(i) Dow’s process
(ii) Finkelstein reaction
(b) Iodobenzene has a higher boiling point than chlorobenzene. Due to the larger size of
the halogen iodine, van der Waals forces are stronger in iodobenzene than
chlorobenzene, making the boiling point of iodobenzene higher than chlorobenzene.
Read More
|
113 videos|348 docs|261 tests
|
About this Document
|
4.96/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: Chemistry: Sample Solution Paper- 4, Class 12 for Grade 12 2024 is part of Chemistry for Grade 12 preparation.
The notes and questions for Chemistry: Sample Solution Paper- 4, Class 12 have been prepared according to the Grade 12 exam syllabus. Information about Chemistry: Sample Solution Paper- 4, Class 12 covers topics
like and Chemistry: Sample Solution Paper- 4, Class 12 Example, for Grade 12 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Chemistry: Sample Solution Paper- 4, Class 12.
Introduction of Chemistry: Sample Solution Paper- 4, Class 12 in English is available as part of our Chemistry for Grade 12
for Grade 12 & Chemistry: Sample Solution Paper- 4, Class 12 in Hindi for Chemistry for Grade 12 course.
Download more important topics related with notes, lectures and mock test series for Grade 12
Exam by signing up for free. Grade 12: Chemistry: Sample Solution Paper- 4, Class 12 | Chemistry for Grade 12
Description
Full syllabus notes, lecture & questions for Chemistry: Sample Solution Paper- 4, Class 12 | Chemistry for Grade 12 - Grade 12 | Plus excerises question with solution to help you revise complete syllabus for Chemistry for Grade 12 | Best notes, free PDF download
Information about Chemistry: Sample Solution Paper- 4, Class 12
In this doc you can find the meaning of Chemistry: Sample Solution Paper- 4, Class 12 defined & explained in the simplest way possible. Besides explaining types of
Chemistry: Sample Solution Paper- 4, Class 12 theory, EduRev gives you an ample number of questions to practice Chemistry: Sample Solution Paper- 4, Class 12 tests, examples and also practice Grade 12
tests
|
Explore Courses for Grade 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
shortcuts and tricks
,Summary
,MCQs
,Chemistry: Sample Solution Paper- 4
,past year papers
,Extra Questions
,Semester Notes
,Class 12 | Chemistry for Grade 12
,Class 12 | Chemistry for Grade 12
,practice quizzes
,ppt
,Important questions
,Previous Year Questions with Solutions
,Objective type Questions
,Free
,video lectures
,study material
,Chemistry: Sample Solution Paper- 4
,mock tests for examination
,Chemistry: Sample Solution Paper- 4
,Class 12 | Chemistry for Grade 12
,Viva Questions
,Sample Paper
,Exam
;
Additional Information about Chemistry: Sample Solution Paper- 4, Class 12 for Grade 12 Preparation
Chemistry: Sample Solution Paper- 4, Class 12 Free PDF Download
The Chemistry: Sample Solution Paper- 4, Class 12 is an invaluable resource that delves deep into the core of the Grade 12 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Chemistry: Sample Solution Paper- 4, Class 12 now and kickstart your journey towards success in the Grade 12 exam.
Importance of Chemistry: Sample Solution Paper- 4, Class 12
The importance of Chemistry: Sample Solution Paper- 4, Class 12 cannot be overstated, especially for Grade 12 aspirants.
This document holds the key to success in the Grade 12 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Chemistry: Sample Solution Paper- 4, Class 12 Notes
Chemistry: Sample Solution Paper- 4, Class 12 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Chemistry: Sample Solution Paper- 4, Class 12.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Chemistry: Sample Solution Paper- 4, Class 12 Notes on EduRev are your ultimate resource for success.
Chemistry: Sample Solution Paper- 4, Class 12 Grade 12 Questions
The "Chemistry: Sample Solution Paper- 4, Class 12 Grade 12 Questions" guide is a valuable resource for all aspiring students preparing for the
Grade 12 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Chemistry: Sample Solution Paper- 4, Class 12 on the App
Students of Grade 12 can study Chemistry: Sample Solution Paper- 4, Class 12 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Chemistry: Sample Solution Paper- 4, Class 12,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Chemistry: Sample Solution Paper- 4, Class 12 is prepared as per the latest Grade 12 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup