Previous Year Paper year:2017 : NEET-XII-Chemistry

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NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 5

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Qstn #61
The biodegradable polymer is:
(1) Nylon-6,6 (2) Nylon-2-Nylon 6
(3) Nylon-6 (4) Buna-S
digAnsr: 2
Ans : ( 2 )
Sol. Nylon-2-Nylon 6

Qstn #62
pH of a saturated solution of ``Ca(OH)_2``
is 9. The
solubility product ``(K_{
sp}) `` of ``Ca(OH)_2`` is:
(1)`` 0.5 × 10^{-15 }
(2) ``0.25 × 10^{-10}``
(3)`` 0.125 × 10^{-15}``
(4)`` 0.5 × 10^{-10}``
digAnsr: 1
Ans : ( 1 )
Sol. + +⇀↽ 22Ca(OH) Ca 2OH
pH = 9 Hence pOH = 14 - 9 = 5
[OH-] = 10-5 M
Hence [Ca2+] =
5
10
2
Thus K
sp
= [Ca2+][OH-]2
=

 
 
 
5
5 210 (10 )
2
= 0.5 × 10-15

Qstn #63
If the rate constant for a first order reaction
is k, the time (t) required for the completion
of 99% of the reaction is given by:
(1) t = 0.693/k
(2) t = 6.909/k
(3) t = 4.606/k
(4) t = 2.303/k
digAnsr: 3
Ans : ( 3 )
Sol. First order rate constant is given as,
= 0
t
[A ]2.303
k log
t [A]
99% completed reaction,
= 2.303 100k log
t 1
= 22.303 log10
t
= 2.303k 2log10
t
=  =2.303 4.606t 2
k k
= 4.606t
k

Qstn #64
The non-essential amino acid among the
following is:
(1) Valine
(2) Leucine
(3) Alanine
(4) Lysine
digAnsr: 3
Ans : ( 3 )
Sol. Alanine
16

Qstn #65
Among the following, the reaction that
proceeds through an electrophilic
substitution, is:
(1)
(2)
(3)
(4)
digAnsr: 2
Ans : ( 2 )
Sol.
Cl
+ Cl
2
AlCl3
+ HCl
Generation of electrophile:
Cl Cl
..
.. .
.
+ AlCl3 Cl Cl AlCl3
+ -
Cl
..
...
. +
AlCl
4
+
-
Electrophile
(i)
+
Cl
+
etc
Cl H
(ii)
+
Cl H
Cl

Qstn #66
The mixture that forms maximum boiling
azeotrope is:
(1) Water + Nitric acid
(2) Ethanol + Water
(3) Acetone + Carbon disulphide
(4) Heptane + Octane
digAnsr: 1
Ans : ( 1 )
Sol.
Answers showing negative deviation from
Raoult's law form maximum boiling azeotrope
Water and Nitric acid  forms maximum
boiling azeotrope

Qstn #67
For the chemical reaction

The correct option is:
(1)
(2)
(3)
(4)
digAnsr: 1
Ans : ( 1 )

Sol. + ⇀↽2 2 3N 3H 2NH
Rate of reaction is given as
 =  = + 32 2
d[NH ]d[N ] d[H ]1 1
dt 3 dt 2 dt

Qstn #68
The number of moles of hydrogen molecules
required to produce 20 moles of ammonia
through Haber's process is :
(1) 10
(2) 20
(3) 30
(4) 40
digAnsr: 3
Ans : ( 3 )
Sol. Haber's process
+
2 2 3
N (g) 3H (g) 2NH (g)
20 moles need to be produced
2 moles of NH
3
 3 moles of H
2
Hence 20 moles of
 =
3
3 20
NH 30
2
moles of H
217

Qstn #69
The compound that is most difficult
to protonate is :
(1)
(2)
(3)
(4)
digAnsr: 4
Ans : ( 4 )
Sol. Due to involvement of lone pair of electrons in
resonance in phenol, it will have positive
charge (partial), hence incoming proton will
not be able to attack easily.

Qstn #70
For an ideal solution, the correct option is :
(1) ``\triangle_{mix}``S = 0 at constant T and P
(2) ``\triangle_{mix}``V ``\neq ``0 at constant T and P
(3)``\triangle_{mix}``H = 0 at constant T and P
(4)``\triangle_{mix}``G = 0 at constant T and P
digAnsr: 3
Ans : ( 3 )
Sol. For ideal solution,
▵
mix
H = 0
▵
mix
S > 0
▵
mix
G < 0
▵
mix
V = 0

Qstn #71
Conjugate base for ``Br\ddot{o}nsted`` acids
``H_2O`` and HF are :
(1)
(2)
(3)
(4)
digAnsr: 3
Ans : ( 3 )
Sol. H O2
H O Conjugate acid
3
+
OH Conjugate base
-
HF on loss of H
+
ion becomes F
-
is the
conjugate base of HF
Example :
HF + H O
2
Acid Base
F + H O
3
+-
Conjugate
base
Conjugate
acid

Qstn #72
Which mixture of the solutions will lead to the
formation of negatively charged colloidal
``[AgI]I^- `` sol ?
(1) 50 mL of 1 M ``AgNO_3`` + 50 mL of 1.5 M KI
(2) 50 mL of 1 M ``AgNO_3`` + 50 mL of 2 M KI
(3) 50 mL of 2 M ``AgNO_3`` + 50 mL of 1.5 M KI
(4) 50 mL of 0.1 M ``AgNO_3`` + 50 mL of 0.1 M KI
digAnsr: 2
Ans : ( 2 )
Sol. Generally charge present on the colloid is due
to adsorption of common ion from dispersion
medium. Millimole of KI is maximum in
option (2) (50 × 2 = 100) so act as solvent and
anion I- is adsorbed by the colloid AgI formed

Qstn #73
Among the following, the narrow spectrum
antibiotic is :
(1) Penicillin G
(2) Ampicillin
(3) Amoxycillin
(4) Chloramphenicol
digAnsr: 1
Ans : ( 1 )
Sol. Penicillin G

Qstn #74
An alkene "A" on reaction with ``O_3``
and ``Zn-H_2O``
gives propanone and ethanal in equimolar
ratio. Addition of HCl to alkene "A" gives "B"
as the major product. The structure of
product "B" is:
(1)
(2)
(3)
(4)
digAnsr: 3
Ans : ( 3 )
18
Sol.

Qstn #75
What is the correct electronic configuration
of the central atom in ``K_4[Fe(CN)_6] ``based on
crystal field theory?
(1)``t_{2g}^4`` ``e_g^2``
(2)``t_{2g}^6`` ``e_g^0``
(3) ``e^3`` ``t_{2}^3``
(4)``e^4`` ``t_{2}^2``
digAnsr: 2
Ans : ( 2 )
Sol. K
4
[Fe(CN)
6
]
Fe ground state: [Ar]3d64s2
Fe2+: 3d64s0
Energy
eg
t2g
In
Spherical
field
6 0
2g gt e
Fe
2+
In the presence of
6CN strong field
-