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# CBSE Chemistry Past year paper (Solutions) - 2015, Class 12 Class 12 Notes | EduRev

## Class 12 : CBSE Chemistry Past year paper (Solutions) - 2015, Class 12 Class 12 Notes | EduRev

``` Page 1

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

CBSE
Class XII Chemistry
Board Paper – 2015 Solution

1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal
particles move towards the oppositely charged electrodes, get discharged and
precipitate.

2. Suppose the number of atoms Y in the hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing,
the number of tetrahedral voids = 2n
As atoms X occupy 2/3
rd
of the tetrahedral voids, the number of atoms X in the lattice

2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?

Hence, the formula of the compound is X4Y3.

3. White phosphorus is less stable and therefore more reactive than the other solid
phases under normal conditions because of the angular strain in the P4 molecule
where the angles are only 60°. Red phosphorus is less reactive than white
phosphorus.

4.

Page 2

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

CBSE
Class XII Chemistry
Board Paper – 2015 Solution

1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal
particles move towards the oppositely charged electrodes, get discharged and
precipitate.

2. Suppose the number of atoms Y in the hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing,
the number of tetrahedral voids = 2n
As atoms X occupy 2/3
rd
of the tetrahedral voids, the number of atoms X in the lattice

2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?

Hence, the formula of the compound is X4Y3.

3. White phosphorus is less stable and therefore more reactive than the other solid
phases under normal conditions because of the angular strain in the P4 molecule
where the angles are only 60°. Red phosphorus is less reactive than white
phosphorus.

4.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the
carbocations, more easily will the product be formed and hence faster will be the rate
of the reaction. Because the stability of the carbocations decreases in the order:
3° carbocation > 2° carbocation > 1° carbocation > CH3
+

Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same
order:
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides
The two structures are

Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1
reaction.
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than
bromoethane.

6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant
temperature is directly proportional to the pressure of the gas present in equilibrium
with the liquid.
Or
The solubility of a gas in a liquid at a particular temperature is directly proportional
to the pressure of the gas in equilibrium with the liquid at that temperature.

The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied
by the evolution of heat. Thus,

Applying Le Chatelier’s principle, the increase of temperature would shift the
equilibrium in the backward direction, that is, solubility would decrease.
Therefore, gases always tend to be less soluble in liquids as the temperature is raised.

OR
Raoult’s law: In a solution, the vapour pressure of a component at a given
temperature is equal to the mole fraction of that component in the solution
multiplied by the vapour pressure of that component in the pure state.

Page 3

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

CBSE
Class XII Chemistry
Board Paper – 2015 Solution

1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal
particles move towards the oppositely charged electrodes, get discharged and
precipitate.

2. Suppose the number of atoms Y in the hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing,
the number of tetrahedral voids = 2n
As atoms X occupy 2/3
rd
of the tetrahedral voids, the number of atoms X in the lattice

2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?

Hence, the formula of the compound is X4Y3.

3. White phosphorus is less stable and therefore more reactive than the other solid
phases under normal conditions because of the angular strain in the P4 molecule
where the angles are only 60°. Red phosphorus is less reactive than white
phosphorus.

4.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the
carbocations, more easily will the product be formed and hence faster will be the rate
of the reaction. Because the stability of the carbocations decreases in the order:
3° carbocation > 2° carbocation > 1° carbocation > CH3
+

Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same
order:
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides
The two structures are

Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1
reaction.
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than
bromoethane.

6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant
temperature is directly proportional to the pressure of the gas present in equilibrium
with the liquid.
Or
The solubility of a gas in a liquid at a particular temperature is directly proportional
to the pressure of the gas in equilibrium with the liquid at that temperature.

The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied
by the evolution of heat. Thus,

Applying Le Chatelier’s principle, the increase of temperature would shift the
equilibrium in the backward direction, that is, solubility would decrease.
Therefore, gases always tend to be less soluble in liquids as the temperature is raised.

OR
Raoult’s law: In a solution, the vapour pressure of a component at a given
temperature is equal to the mole fraction of that component in the solution
multiplied by the vapour pressure of that component in the pure state.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

Ideal Solution Non-ideal Solution
1. An ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of the same
magnitude as the
intermolecular interactions
found in the pure components
(A–A attractions and B–B
attractions).
1. A non-ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of different
magnitude as the intermolecular
interactions found in the pure
components (A–A attractions
and B–B attractions).
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0

7.
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution:

0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2

The reaction with more positive value of reduction potential would undergo
reduction at the cathode.
Here, in given reactions, reduction potential of H
+
is more positive than reduction
potential of Na
+
. The reduction potential value of Na
+
is negative.
So, the reduction of H
+
is feasible at the cathode.

(b)
? A mercury cell consists of a zinc container as the anode, a carbon rod as the
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.
? In this cell, the overall cell reaction does not involve any ion whose
concentration may change.
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.

8. Transition metals have high effective nuclear charge, greater number of valence
electrons and some unpaired electrons. They thus have strong metal–metal bonding.
Hence, transition metals have high enthalpies of atomisation.
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc.
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition
elements.
Page 4

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

CBSE
Class XII Chemistry
Board Paper – 2015 Solution

1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal
particles move towards the oppositely charged electrodes, get discharged and
precipitate.

2. Suppose the number of atoms Y in the hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing,
the number of tetrahedral voids = 2n
As atoms X occupy 2/3
rd
of the tetrahedral voids, the number of atoms X in the lattice

2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?

Hence, the formula of the compound is X4Y3.

3. White phosphorus is less stable and therefore more reactive than the other solid
phases under normal conditions because of the angular strain in the P4 molecule
where the angles are only 60°. Red phosphorus is less reactive than white
phosphorus.

4.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the
carbocations, more easily will the product be formed and hence faster will be the rate
of the reaction. Because the stability of the carbocations decreases in the order:
3° carbocation > 2° carbocation > 1° carbocation > CH3
+

Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same
order:
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides
The two structures are

Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1
reaction.
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than
bromoethane.

6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant
temperature is directly proportional to the pressure of the gas present in equilibrium
with the liquid.
Or
The solubility of a gas in a liquid at a particular temperature is directly proportional
to the pressure of the gas in equilibrium with the liquid at that temperature.

The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied
by the evolution of heat. Thus,

Applying Le Chatelier’s principle, the increase of temperature would shift the
equilibrium in the backward direction, that is, solubility would decrease.
Therefore, gases always tend to be less soluble in liquids as the temperature is raised.

OR
Raoult’s law: In a solution, the vapour pressure of a component at a given
temperature is equal to the mole fraction of that component in the solution
multiplied by the vapour pressure of that component in the pure state.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

Ideal Solution Non-ideal Solution
1. An ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of the same
magnitude as the
intermolecular interactions
found in the pure components
(A–A attractions and B–B
attractions).
1. A non-ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of different
magnitude as the intermolecular
interactions found in the pure
components (A–A attractions
and B–B attractions).
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0

7.
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution:

0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2

The reaction with more positive value of reduction potential would undergo
reduction at the cathode.
Here, in given reactions, reduction potential of H
+
is more positive than reduction
potential of Na
+
. The reduction potential value of Na
+
is negative.
So, the reduction of H
+
is feasible at the cathode.

(b)
? A mercury cell consists of a zinc container as the anode, a carbon rod as the
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.
? In this cell, the overall cell reaction does not involve any ion whose
concentration may change.
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.

8. Transition metals have high effective nuclear charge, greater number of valence
electrons and some unpaired electrons. They thus have strong metal–metal bonding.
Hence, transition metals have high enthalpies of atomisation.
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc.
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition
elements.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

9.
(i) [Co(NH3)5(NO2)](NO3)2
IUPAC name: Pentaamminenitrocobalt (III) nitrate
(ii) Potassium tetracyanidonickelate (II)
Formula of the complex: K2[Ni(CN)4]

10.
(i)

(ii)

11. Given:
Molar mass of CaCl2 (MB)  = 111 g/mol
Weight of water (wA)  = 500 g
Kf for water   = 1.86 K kg/mol
?Tf    = 2 K
Formula:
fB
f
AB
K × w × 1000
? T =
w × M

Solution:
fB
f
AB
B
B
K × w × 1000
? T =
w × M
1.86 × w × 1000
2=
500 × 111
2 × 500 × 111
w = = 59.68
1.86 × 1000

Amount of CaCl2 required = 59.68 g

Page 5

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

CBSE
Class XII Chemistry
Board Paper – 2015 Solution

1. Lyophobic colloids can be coagulated by the electrophoresis method. The colloidal
particles move towards the oppositely charged electrodes, get discharged and
precipitate.

2. Suppose the number of atoms Y in the hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing,
the number of tetrahedral voids = 2n
As atoms X occupy 2/3
rd
of the tetrahedral voids, the number of atoms X in the lattice

2 4n
×2n =
33
4n 4
Ratioof X:Y = :n = :1= 4:3
33
?

Hence, the formula of the compound is X4Y3.

3. White phosphorus is less stable and therefore more reactive than the other solid
phases under normal conditions because of the angular strain in the P4 molecule
where the angles are only 60°. Red phosphorus is less reactive than white
phosphorus.

4.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

5. Carbocations are the intermediates in the SN1 reaction. Greater the stability of the
carbocations, more easily will the product be formed and hence faster will be the rate
of the reaction. Because the stability of the carbocations decreases in the order:
3° carbocation > 2° carbocation > 1° carbocation > CH3
+

Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same
order:
3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides
The two structures are

Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in
the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a
tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1
reaction.
Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than
bromoethane.

6. Henry’s law: The mass of a gas dissolved in a given volume of the liquid at constant
temperature is directly proportional to the pressure of the gas present in equilibrium
with the liquid.
Or
The solubility of a gas in a liquid at a particular temperature is directly proportional
to the pressure of the gas in equilibrium with the liquid at that temperature.

The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied
by the evolution of heat. Thus,

Applying Le Chatelier’s principle, the increase of temperature would shift the
equilibrium in the backward direction, that is, solubility would decrease.
Therefore, gases always tend to be less soluble in liquids as the temperature is raised.

OR
Raoult’s law: In a solution, the vapour pressure of a component at a given
temperature is equal to the mole fraction of that component in the solution
multiplied by the vapour pressure of that component in the pure state.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

Ideal Solution Non-ideal Solution
1. An ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of the same
magnitude as the
intermolecular interactions
found in the pure components
(A–A attractions and B–B
attractions).
1. A non-ideal solution of the
components A and B is the
solution in which the
intermolecular interactions
between the components (A–B
attractions) are of different
magnitude as the intermolecular
interactions found in the pure
components (A–A attractions
and B–B attractions).
2. ?mixH = 0 and ?mixV = 0 2. ?mixH ? 0 and ?mixV ? 0

7.
(a) Reactions taking place during the electrolysis of aqueous sodium chloride solution:

0.00
?
?
+ - o
+ - o
2
Na (aq)+e Na(s) E =-2.71 V (I)
1
H (aq)+ e H (g) E = V (II)
2

The reaction with more positive value of reduction potential would undergo
reduction at the cathode.
Here, in given reactions, reduction potential of H
+
is more positive than reduction
potential of Na
+
. The reduction potential value of Na
+
is negative.
So, the reduction of H
+
is feasible at the cathode.

(b)
? A mercury cell consists of a zinc container as the anode, a carbon rod as the
cathode and a paste of mercuric oxide mixed with KOH as the electrolyte.
? In this cell, the overall cell reaction does not involve any ion whose
concentration may change.
? Therefore, this cell gives a constant potential of 1.5 V throughout its life.

8. Transition metals have high effective nuclear charge, greater number of valence
electrons and some unpaired electrons. They thus have strong metal–metal bonding.
Hence, transition metals have high enthalpies of atomisation.
In the 3d series, from Sc to Zn, only zinc has filled valence shells. The valence shell
electronic configuration of Zn is 3d
10
4s
2
. Because of the absence of unpaired electrons
in ns and (n–1) d shells, the interatomic electronic bonding is the weakest in zinc.
Consequently, zinc has the least enthalpy of atomisation in the 3d series of transition
elements.

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

9.
(i) [Co(NH3)5(NO2)](NO3)2
IUPAC name: Pentaamminenitrocobalt (III) nitrate
(ii) Potassium tetracyanidonickelate (II)
Formula of the complex: K2[Ni(CN)4]

10.
(i)

(ii)

11. Given:
Molar mass of CaCl2 (MB)  = 111 g/mol
Weight of water (wA)  = 500 g
Kf for water   = 1.86 K kg/mol
?Tf    = 2 K
Formula:
fB
f
AB
K × w × 1000
? T =
w × M

Solution:
fB
f
AB
B
B
K × w × 1000
? T =
w × M
1.86 × w × 1000
2=
500 × 111
2 × 500 × 111
w = = 59.68
1.86 × 1000

Amount of CaCl2 required = 59.68 g

CBSE XII  | CHEMISTRY
Board Paper – 2015 Solution

12. Given:
Density (D)   =  10 g/cm
3

Edge length (a)  =  3 × 10
-8
cm
Atomic mass   =  81 g/mol
Formula: 3
0
Z × M
Density =
a × N

3
0
3 23
3 23
Z × M
Density =
a × N
Z × 81
10 =
(3 10) × 6.022 × 10
10 × (3 10) × 6.022 × 10
Z = = 2.0073
81
?
?

? Z = 2
The unit cell contains 2 atoms, so it is a body-centred cubic unit cell.

13. According to Nernst equation,

0
cell cell
0.0592 [P]
E = E log
n [R]
?

2+ 2+
2
0 0 0
cell Oxi(anode) Oxi(cathode)
00
Sn|Sn H |2H
E E E
EE
0.14 0.0
0.14
??
??
? ? ?
??

Reactions:
Anode (oxidation):   Sn(s) ? Sn2
+
(aq) + 2e
-

Cathode (reduction):   2H
+
(aq) + 2e
-
? H2(g)
Net reaction:    Sn(s) + 2H
+
(aq) ? Sn2
+
(aq) + H2(g)
? n = 2
[P] = [Sn2
+
] = 0.001 M
[R] = [H
+
] = 0.01 M
So, putting the above values in the formula,
cell 2
0.0592 [0.001]
E = 0.14 log
2 [0.01]
??

Ecell = –0.1696 V

```
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