Chemistry: Sample Solution Paper- 3, Class 12 Notes | Study Chemistry Class 12 - NEET
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Page 1
CBSE XII | Chemistry
Sample Paper 3 - Solution
CBSE
Class XII Chemistry
Sample Paper 3 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. A polymer formed by the polymerisation of two or more different monomers is called a
copolymer; for example, Buna-S.
2. This is because amorphous solids have a tendency to flow.
OR
Frenkel defect is shown by ionic compounds having large difference in the size of ions
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy
interstitial sites.
3. Pyridinium chlorochromate
4. Yes, if one of the reactants is taken in excess, its concentration does not change
considerably with time. So, a bimolecular reaction can be of the first order.
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be
used for drying ammonia.
OR
The increasing order of acidic character of haloacids is
HF < HCl < HBr < HI
Section B
6.
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the
number of moles of that component to the total number of moles of all the
components of the mixture.
Mole fraction is given by
nA =
Molesof A
Total,molesof allthecomponentsof themixtur e
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
OR
Page 2
CBSE XII | Chemistry
Sample Paper 3 - Solution
CBSE
Class XII Chemistry
Sample Paper 3 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. A polymer formed by the polymerisation of two or more different monomers is called a
copolymer; for example, Buna-S.
2. This is because amorphous solids have a tendency to flow.
OR
Frenkel defect is shown by ionic compounds having large difference in the size of ions
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy
interstitial sites.
3. Pyridinium chlorochromate
4. Yes, if one of the reactants is taken in excess, its concentration does not change
considerably with time. So, a bimolecular reaction can be of the first order.
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be
used for drying ammonia.
OR
The increasing order of acidic character of haloacids is
HF < HCl < HBr < HI
Section B
6.
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the
number of moles of that component to the total number of moles of all the
components of the mixture.
Mole fraction is given by
nA =
Molesof A
Total,molesof allthecomponentsof themixtur e
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
OR
CBSE XII | Chemistry
Sample Paper 3 - Solution
The relationship between mole fraction and vapour pressure of a component of
an ideal solution in the liquid phase and vapour phase:
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
7.
When Kc < 1, taking log, gives a negative value.
For example:
Thus, is negative if the equilibrium constant Kc < 1.
8. Given:
Rate constant K = 200 s
-1
The half-life of the first-order reaction is
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ?
9.
(a) Cl2O:
2x – 2 = 0
2x = 2
x = + 1
(b) KBrO3:
1 + x - 6 = 0
x = 6 + 1
x = 7
cell c
0.059
For a cell, E = log K
n
?
cell
0.059
E = log 0.01
n
= - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1
?
?
?
?
?
Page 3
CBSE XII | Chemistry
Sample Paper 3 - Solution
CBSE
Class XII Chemistry
Sample Paper 3 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. A polymer formed by the polymerisation of two or more different monomers is called a
copolymer; for example, Buna-S.
2. This is because amorphous solids have a tendency to flow.
OR
Frenkel defect is shown by ionic compounds having large difference in the size of ions
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy
interstitial sites.
3. Pyridinium chlorochromate
4. Yes, if one of the reactants is taken in excess, its concentration does not change
considerably with time. So, a bimolecular reaction can be of the first order.
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be
used for drying ammonia.
OR
The increasing order of acidic character of haloacids is
HF < HCl < HBr < HI
Section B
6.
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the
number of moles of that component to the total number of moles of all the
components of the mixture.
Mole fraction is given by
nA =
Molesof A
Total,molesof allthecomponentsof themixtur e
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
OR
CBSE XII | Chemistry
Sample Paper 3 - Solution
The relationship between mole fraction and vapour pressure of a component of
an ideal solution in the liquid phase and vapour phase:
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
7.
When Kc < 1, taking log, gives a negative value.
For example:
Thus, is negative if the equilibrium constant Kc < 1.
8. Given:
Rate constant K = 200 s
-1
The half-life of the first-order reaction is
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ?
9.
(a) Cl2O:
2x – 2 = 0
2x = 2
x = + 1
(b) KBrO3:
1 + x - 6 = 0
x = 6 + 1
x = 7
cell c
0.059
For a cell, E = log K
n
?
cell
0.059
E = log 0.01
n
= - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1
?
?
?
?
?
CBSE XII | Chemistry
Sample Paper 3 - Solution
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of
iodoform, while methanol will not.
C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa
Yellow
CH3OH + I2 + NaOH no yellow ppt.
11.
(a) Methyl-2-aminobutanoate
(b) N-methylbenzenamine
OR
(a) Ambident nucleophile
(i)
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ?
(ii)
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ?
From the above reactions, it is clear that CN
-
is an ambident nucleophile.
(b) Hinsberg test:
?
?
Page 4
CBSE XII | Chemistry
Sample Paper 3 - Solution
CBSE
Class XII Chemistry
Sample Paper 3 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. A polymer formed by the polymerisation of two or more different monomers is called a
copolymer; for example, Buna-S.
2. This is because amorphous solids have a tendency to flow.
OR
Frenkel defect is shown by ionic compounds having large difference in the size of ions
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy
interstitial sites.
3. Pyridinium chlorochromate
4. Yes, if one of the reactants is taken in excess, its concentration does not change
considerably with time. So, a bimolecular reaction can be of the first order.
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be
used for drying ammonia.
OR
The increasing order of acidic character of haloacids is
HF < HCl < HBr < HI
Section B
6.
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the
number of moles of that component to the total number of moles of all the
components of the mixture.
Mole fraction is given by
nA =
Molesof A
Total,molesof allthecomponentsof themixtur e
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
OR
CBSE XII | Chemistry
Sample Paper 3 - Solution
The relationship between mole fraction and vapour pressure of a component of
an ideal solution in the liquid phase and vapour phase:
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
7.
When Kc < 1, taking log, gives a negative value.
For example:
Thus, is negative if the equilibrium constant Kc < 1.
8. Given:
Rate constant K = 200 s
-1
The half-life of the first-order reaction is
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ?
9.
(a) Cl2O:
2x – 2 = 0
2x = 2
x = + 1
(b) KBrO3:
1 + x - 6 = 0
x = 6 + 1
x = 7
cell c
0.059
For a cell, E = log K
n
?
cell
0.059
E = log 0.01
n
= - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1
?
?
?
?
?
CBSE XII | Chemistry
Sample Paper 3 - Solution
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of
iodoform, while methanol will not.
C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa
Yellow
CH3OH + I2 + NaOH no yellow ppt.
11.
(a) Methyl-2-aminobutanoate
(b) N-methylbenzenamine
OR
(a) Ambident nucleophile
(i)
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ?
(ii)
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ?
From the above reactions, it is clear that CN
-
is an ambident nucleophile.
(b) Hinsberg test:
?
?
CBSE XII | Chemistry
Sample Paper 3 - Solution
12.
(a) Benzoyl peroxide acts as an initiator because it generates free radicals.
(b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene.
LDPE: It is produced by free radical polymerisation at a high temperature of 350–
570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer.
HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta
catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a
linear polymer.
Section C
13. Given:
Cell edge (a) = 288 pm = 288 ×10
-10
cm
Let’s first find the volume,
Volume of unit cell = (288 ×10
-10
cm)
3
= 2.389×10
-23
cm
3
3
Mass
Volume of 208 g of the element
Density
208
7.2
28.89cm
?
?
?
23
23 3
Totalvolume
Numberof unit cells
Volumeof aunit cell
28.89
2.389 10
12.09 10 cm
?
?
?
?
?
??
For a b.c.c. structure, the number of atoms per unit cell = 2
Number of atoms present in 208 g
= No. of atoms per unit cell × No. of unit cells
= 2 × 12.09 × 10
23
= 24.18 × 10
23
= 2.418×10
24
atoms
Page 5
CBSE XII | Chemistry
Sample Paper 3 - Solution
CBSE
Class XII Chemistry
Sample Paper 3 - Solution
Time: 3 Hrs Maximum Marks: 70
_________________________________________________________________________________________________________
Section A
1. A polymer formed by the polymerisation of two or more different monomers is called a
copolymer; for example, Buna-S.
2. This is because amorphous solids have a tendency to flow.
OR
Frenkel defect is shown by ionic compounds having large difference in the size of ions
so that the smaller ion dislocates from its normal site and occupies an interstitial site. In
case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy
interstitial sites.
3. Pyridinium chlorochromate
4. Yes, if one of the reactants is taken in excess, its concentration does not change
considerably with time. So, a bimolecular reaction can be of the first order.
5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be
used for drying ammonia.
OR
The increasing order of acidic character of haloacids is
HF < HCl < HBr < HI
Section B
6.
(a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the
number of moles of that component to the total number of moles of all the
components of the mixture.
Mole fraction is given by
nA =
Molesof A
Total,molesof allthecomponentsof themixtur e
(b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as
Observedvalueof colligativeproperty
i=
Normalvalueof colligativeproperty
OR
CBSE XII | Chemistry
Sample Paper 3 - Solution
The relationship between mole fraction and vapour pressure of a component of
an ideal solution in the liquid phase and vapour phase:
0
A A A
0
B B B
AB
00
A A B B
P =x .P
P =x .P
P =P +P
= x .P +x .P
7.
When Kc < 1, taking log, gives a negative value.
For example:
Thus, is negative if the equilibrium constant Kc < 1.
8. Given:
Rate constant K = 200 s
-1
The half-life of the first-order reaction is
3
1
2
0.693 0.693
t 3.46 10 sec.
K 200
?
? ? ? ?
9.
(a) Cl2O:
2x – 2 = 0
2x = 2
x = + 1
(b) KBrO3:
1 + x - 6 = 0
x = 6 + 1
x = 7
cell c
0.059
For a cell, E = log K
n
?
cell
0.059
E = log 0.01
n
= - 2 x 0.059/n ( negative value)
?
cell
E
?
cell
c
c
c
c
If E 0
0.059
Then 0 = log K
n
log K 0
K Antilog(0)
K = 1
?
?
?
?
?
CBSE XII | Chemistry
Sample Paper 3 - Solution
10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of
iodoform, while methanol will not.
C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa
Yellow
CH3OH + I2 + NaOH no yellow ppt.
11.
(a) Methyl-2-aminobutanoate
(b) N-methylbenzenamine
OR
(a) Ambident nucleophile
(i)
2 5 2 5
C H Cl+KCN C H CN+KCl ? ? ?
(ii)
2 5 2 5
C H Cl+AgCN C H -N C+AgCl ? ? ? ?
From the above reactions, it is clear that CN
-
is an ambident nucleophile.
(b) Hinsberg test:
?
?
CBSE XII | Chemistry
Sample Paper 3 - Solution
12.
(a) Benzoyl peroxide acts as an initiator because it generates free radicals.
(b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene.
LDPE: It is produced by free radical polymerisation at a high temperature of 350–
570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer.
HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta
catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a
linear polymer.
Section C
13. Given:
Cell edge (a) = 288 pm = 288 ×10
-10
cm
Let’s first find the volume,
Volume of unit cell = (288 ×10
-10
cm)
3
= 2.389×10
-23
cm
3
3
Mass
Volume of 208 g of the element
Density
208
7.2
28.89cm
?
?
?
23
23 3
Totalvolume
Numberof unit cells
Volumeof aunit cell
28.89
2.389 10
12.09 10 cm
?
?
?
?
?
??
For a b.c.c. structure, the number of atoms per unit cell = 2
Number of atoms present in 208 g
= No. of atoms per unit cell × No. of unit cells
= 2 × 12.09 × 10
23
= 24.18 × 10
23
= 2.418×10
24
atoms
CBSE XII | Chemistry
Sample Paper 3 - Solution
14. Given:
1
1
1
2
1
1
1
0
1 1 2 2 1
0
1 1 2 2 1
0
11
0
1
1
0
1
1
0
1
1
0
1
1
1
W 90g
W 10g
M 72 12 96 180g mol
P 32.8mmHg
W 18g mol
P P n W M
P n n M W
PP 92 18
P 180 90
P 92 18
1
P 180 90
P 92 18
1
P 180 90
P 92 18
1
P 180 90
P 92 18
1
32.8 180 90
P 0.36KPa
?
?
?
?
? ? ? ?
?
?
?
? ? ?
?
?
??
? ? ?
?
? ? ?
?
? ??
??
??
?
??
?
??
?
?
OR
8 18
o
oo
s
B
B
1
CH 1
80 P
P 80%of P 0.80P
100
W
n mol
40
114g
n 1mol...................(Molar mass 114gmol)
114gmol
?
?
?
? ? ?
? ? ?
8 18
B
os
B
B o
B
B C H
w
n pp
40
Molesof octane x
W
p n n
40 1
?
? ? ? ? ?
?
?
B
oo
o
B
B
B
w
p 0.80P
40
W
P
40 1
w
W
40
1 0.80
W
W 40
40 1
?
?
?
? ? ?
?
?
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