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# Chemistry: Sample Solution Paper- 4, Class 12 Notes | EduRev

## Class 12 : Chemistry: Sample Solution Paper- 4, Class 12 Notes | EduRev

``` 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.
```
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