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# Sample Solution Paper 5 - Chemistry, Class 12 NEET Notes | EduRev

## NEET : Sample Solution Paper 5 - Chemistry, Class 12 NEET Notes | EduRev

``` Page 1

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

CBSE
Class  XII Chemistry
Sample  Paper 5 - Solution

Time: 3 Hrs                                                                Total Marks: 70

Section A

1. Amorphous solid
OR

The coordination number of hcp and ccp is 12.

2. A finely divided substance provides more surface area for adsorption and hence is more
effective as an absorbent.

3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for
lighters.
OR

KMnO4 has +7 oxidation state.

4. Cyclopentane carbaldehyde

5. Phenol and formaldehyde

Section B
6. Molality

222.6
= = 17.95m
62× 0.2

Mass of solution = 200 + 222.6 = 422.6 g

-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL

Molarity
B
B
w 1000
=×
M Volumeof solution(in mL)

222.6 1000
=×
62 394.22
=9.1M

OR

Page 2

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

CBSE
Class  XII Chemistry
Sample  Paper 5 - Solution

Time: 3 Hrs                                                                Total Marks: 70

Section A

1. Amorphous solid
OR

The coordination number of hcp and ccp is 12.

2. A finely divided substance provides more surface area for adsorption and hence is more
effective as an absorbent.

3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for
lighters.
OR

KMnO4 has +7 oxidation state.

4. Cyclopentane carbaldehyde

5. Phenol and formaldehyde

Section B
6. Molality

222.6
= = 17.95m
62× 0.2

Mass of solution = 200 + 222.6 = 422.6 g

-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL

Molarity
B
B
w 1000
=×
M Volumeof solution(in mL)

222.6 1000
=×
62 394.22
=9.1M

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g
mol
-1
.
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?

7.
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??

8.
(a)
1
s
?

(b) Slope = -k

9.
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-

ion is known, but phosphorus in CP
-
ion does not form ???? - ???? bond due to its
bigger size.
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore
gets dimerised to the stable N2O4.
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2.

10.
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to
form anilinium ion, which is a meta directing group, thus giving substantial amount
of m-nitroaniline.

(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its
pKb value will be higher. On the other hand, the +I effect of the
3
-CH group,
increases the electron density on the N atom in
32
CH NH making it a stronger base.
Hence, its Kb value will be higher and its pKb value will be lower.

OR
Page 3

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

CBSE
Class  XII Chemistry
Sample  Paper 5 - Solution

Time: 3 Hrs                                                                Total Marks: 70

Section A

1. Amorphous solid
OR

The coordination number of hcp and ccp is 12.

2. A finely divided substance provides more surface area for adsorption and hence is more
effective as an absorbent.

3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for
lighters.
OR

KMnO4 has +7 oxidation state.

4. Cyclopentane carbaldehyde

5. Phenol and formaldehyde

Section B
6. Molality

222.6
= = 17.95m
62× 0.2

Mass of solution = 200 + 222.6 = 422.6 g

-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL

Molarity
B
B
w 1000
=×
M Volumeof solution(in mL)

222.6 1000
=×
62 394.22
=9.1M

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g
mol
-1
.
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?

7.
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??

8.
(a)
1
s
?

(b) Slope = -k

9.
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-

ion is known, but phosphorus in CP
-
ion does not form ???? - ???? bond due to its
bigger size.
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore
gets dimerised to the stable N2O4.
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2.

10.
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to
form anilinium ion, which is a meta directing group, thus giving substantial amount
of m-nitroaniline.

(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its
pKb value will be higher. On the other hand, the +I effect of the
3
-CH group,
increases the electron density on the N atom in
32
CH NH making it a stronger base.
Hence, its Kb value will be higher and its pKb value will be lower.

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding.
Thus, primary amines have a higher boiling point than tertiary amines as a result of
their hydrogen bonding.

(b)
+

11.
(a) Butanone < Propanone < Propanal < Ethanal
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde

12.
(a) Polymers which are degraded by microorganisms within a suitable period so that
the polymers and their degraded products do not cause any serious effects on the
Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)

(b) On the basis of molecular forces present between the chains of various polymers,
polymers are classified as
1. Elastomers
2. Fibres
3. Thermoplastics
4. Thermosetting plastics

Section C

13.
(a) The co-ordination number of each sphere in the bcc structure is 8.
(b) Number of A atoms =
1
8x
8
= 1 atom
Number of B atoms =
1
6x
2
= 3 atoms
Therefore, the formula of the crystalline compound = AB3
(c) The compound of group 12â€“16 is zinc sulphide.

Page 4

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

CBSE
Class  XII Chemistry
Sample  Paper 5 - Solution

Time: 3 Hrs                                                                Total Marks: 70

Section A

1. Amorphous solid
OR

The coordination number of hcp and ccp is 12.

2. A finely divided substance provides more surface area for adsorption and hence is more
effective as an absorbent.

3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for
lighters.
OR

KMnO4 has +7 oxidation state.

4. Cyclopentane carbaldehyde

5. Phenol and formaldehyde

Section B
6. Molality

222.6
= = 17.95m
62× 0.2

Mass of solution = 200 + 222.6 = 422.6 g

-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL

Molarity
B
B
w 1000
=×
M Volumeof solution(in mL)

222.6 1000
=×
62 394.22
=9.1M

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g
mol
-1
.
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?

7.
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??

8.
(a)
1
s
?

(b) Slope = -k

9.
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-

ion is known, but phosphorus in CP
-
ion does not form ???? - ???? bond due to its
bigger size.
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore
gets dimerised to the stable N2O4.
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2.

10.
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to
form anilinium ion, which is a meta directing group, thus giving substantial amount
of m-nitroaniline.

(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its
pKb value will be higher. On the other hand, the +I effect of the
3
-CH group,
increases the electron density on the N atom in
32
CH NH making it a stronger base.
Hence, its Kb value will be higher and its pKb value will be lower.

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding.
Thus, primary amines have a higher boiling point than tertiary amines as a result of
their hydrogen bonding.

(b)
+

11.
(a) Butanone < Propanone < Propanal < Ethanal
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde

12.
(a) Polymers which are degraded by microorganisms within a suitable period so that
the polymers and their degraded products do not cause any serious effects on the
Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)

(b) On the basis of molecular forces present between the chains of various polymers,
polymers are classified as
1. Elastomers
2. Fibres
3. Thermoplastics
4. Thermosetting plastics

Section C

13.
(a) The co-ordination number of each sphere in the bcc structure is 8.
(b) Number of A atoms =
1
8x
8
= 1 atom
Number of B atoms =
1
6x
2
= 3 atoms
Therefore, the formula of the crystalline compound = AB3
(c) The compound of group 12â€“16 is zinc sulphide.

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

14.
?
ff
B
f
BA
fB
B
fA
?T = K m
w 1000
= K ×
Mw
K × 1000 x w
M=
?T ×w
4.9 × 1000 × 2
=
1.62 × 25
-1
= 241.98 g mol

Normal molar mass of benzoic acid = 122 g mol
-1
i =
NormalMolarMass 122
= =0.504
ObservedMolar Mass 241.98

? ? ? 2 C H COOH (C H COOH)
6 6 2 55
Initial 1 0
After
a
association 1 - a
2
Therefore,
a
1- a+
2
i=
1
a
i=1 -
2
a
i=1 -
2
2i=2- a
a = 2 - 2i
a = 2 - 2 x 0. 50 4
=2 -1.008
=0.992
a = 99 .2 %

Page 5

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

CBSE
Class  XII Chemistry
Sample  Paper 5 - Solution

Time: 3 Hrs                                                                Total Marks: 70

Section A

1. Amorphous solid
OR

The coordination number of hcp and ccp is 12.

2. A finely divided substance provides more surface area for adsorption and hence is more
effective as an absorbent.

3. Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for
lighters.
OR

KMnO4 has +7 oxidation state.

4. Cyclopentane carbaldehyde

5. Phenol and formaldehyde

Section B
6. Molality

222.6
= = 17.95m
62× 0.2

Mass of solution = 200 + 222.6 = 422.6 g

-1
Mass of solution
Density of solution =
Volume of solution
422.6 g
1.072g mL =
Volume of solution
Volume of solution =394.22 mL

Molarity
B
B
w 1000
=×
M Volumeof solution(in mL)

222.6 1000
=×
62 394.22
=9.1M

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

The molecular mass of ascorbic acid (C6H8O6) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g
mol
-1
.
B
ff
BA
f B A
B
f
1
B 1
1000 W
TK
MW
T M W
W
K 1000
1.5K 176gmol 75g
W 5.08g
3.9K kg mol 1000
?
?
?
??
?
? ? ?
?
?
??
??
?

7.
0 0 0 2
cell 1/2 2
0
34
E E Cl / Cl E Cu / Cu
1.36V 0.34V 1.02V
nE 2 1.02
log k
0.0591 0.0591
2.04
34.5177
0.0591
k antilog34.5177
k 3.294 10
??
??
? ? ? ?
?
??
??
?
??

8.
(a)
1
s
?

(b) Slope = -k

9.
(a) Nitrogen being smaller in size forms ???? - ???? multiple bonding with carbon, so CN
-

ion is known, but phosphorus in CP
-
ion does not form ???? - ???? bond due to its
bigger size.
(b) NO2 dimerises to form N2O5 because NO2 is an odd electron molecule and therefore
gets dimerised to the stable N2O4.
(c) ICl is more reactive than I2 because ICl has less bond dissociation enthalpy than I2.

10.
(a) In the presence of nitrating mixture ? ?
3 2 4
HNO +H SO , aniline gets protonated to
form anilinium ion, which is a meta directing group, thus giving substantial amount
of m-nitroaniline.

(b) In aniline, a lone pair of electrons on the N atom is delocalised over the benzene
ring, resulting in lowering its basic strength. Hence, its Kb value will be lower and its
pKb value will be higher. On the other hand, the +I effect of the
3
-CH group,
increases the electron density on the N atom in
32
CH NH making it a stronger base.
Hence, its Kb value will be higher and its pKb value will be lower.

OR

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

(a) Primary amines (RNH2) have two hydrogen atoms attached to the nitrogen atom
and therefore form hydrogen bonding. Tertiary amines (R3N) do not have hydrogen
atoms attached to the nitrogen atom and therefore do not show hydrogen bonding.
Thus, primary amines have a higher boiling point than tertiary amines as a result of
their hydrogen bonding.

(b)
+

11.
(a) Butanone < Propanone < Propanal < Ethanal
(b) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde

12.
(a) Polymers which are degraded by microorganisms within a suitable period so that
the polymers and their degraded products do not cause any serious effects on the
Example: Poly- ?-hydroxybutyrate-co- ?-hydroxy valerate (PHBV)

(b) On the basis of molecular forces present between the chains of various polymers,
polymers are classified as
1. Elastomers
2. Fibres
3. Thermoplastics
4. Thermosetting plastics

Section C

13.
(a) The co-ordination number of each sphere in the bcc structure is 8.
(b) Number of A atoms =
1
8x
8
= 1 atom
Number of B atoms =
1
6x
2
= 3 atoms
Therefore, the formula of the crystalline compound = AB3
(c) The compound of group 12â€“16 is zinc sulphide.

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

14.
?
ff
B
f
BA
fB
B
fA
?T = K m
w 1000
= K ×
Mw
K × 1000 x w
M=
?T ×w
4.9 × 1000 × 2
=
1.62 × 25
-1
= 241.98 g mol

Normal molar mass of benzoic acid = 122 g mol
-1
i =
NormalMolarMass 122
= =0.504
ObservedMolar Mass 241.98

? ? ? 2 C H COOH (C H COOH)
6 6 2 55
Initial 1 0
After
a
association 1 - a
2
Therefore,
a
1- a+
2
i=
1
a
i=1 -
2
a
i=1 -
2
2i=2- a
a = 2 - 2i
a = 2 - 2 x 0. 50 4
=2 -1.008
=0.992
a = 99 .2 %

CBSE XII  |  Chemistry
Sample Paper  5 - Solution

15.    For a first-order reaction:

2.303 a
k = log
t a - x

-5
2.303 a
2.3 × 10 = log
200 × 60 a - x
a
log =0.1198
a - x
a
= Antilog (0.1198)
a - x
a
=1.317
a - x
a = 1.317(a - x)
1 a - x
=
1.317 a
x
1 - =0.759
a
x
=1-0.759=0.241
a

x
= 0.241
a

% decomposed = 24.1%
% remained = 100 - 24.1 = 75.9%
OR

16.
(a) Alcosol: A colloidal sol in which the dispersion medium is alcohol. Example:
Collodion
(b) Aerosol: When the dispersion medium is a gas and the dispersed phase is either
solid or liquid, the colloidal system is called an aerosol. Examples: Fog, cloud, smoke
(c) Hydrosol: Colloids in water are called hydrosols. Examples: Milk, protein

```
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