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Page 1
CHEMISTRY
Q 1. Following limiting molar conductivities are given as
?° m (H 2SO 4) = x S cm
2
mol
-1
?° m (K 2SO 4) = y S cm
2
mol
-1
?° m (CH 3COOK) = z S cm
2
mol
-1
?° m (in S cm
2
mol
-1
) for CH 3COOH will be
Option A
(x y)
z
2
?
?
Option B x – y + 2z
Option C x + y + z
Option D x - y + z
Correct Option A
Solution:
Molar conductivity of AB is calculated as-
Page 2
CHEMISTRY
Q 1. Following limiting molar conductivities are given as
?° m (H 2SO 4) = x S cm
2
mol
-1
?° m (K 2SO 4) = y S cm
2
mol
-1
?° m (CH 3COOK) = z S cm
2
mol
-1
?° m (in S cm
2
mol
-1
) for CH 3COOH will be
Option A
(x y)
z
2
?
?
Option B x – y + 2z
Option C x + y + z
Option D x - y + z
Correct Option A
Solution:
Molar conductivity of AB is calculated as-
? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ?
m m m
m 3 m 3 m
m3
m 3 m 2 4 m 2 4
AB A B
So , CH COOH CH COO H
So CH COOH
11
CH COOK H SO K SO
22
xy
z
22
xy
z
2
??
??
? ? ? ? ?
? ? ? ? ?
?
? ? ? ? ? ?
? ? ?
? ??
??
??
??
Q 2. A first order reaction has a rate constant of 2.303 × 10
–3
s
–1
. The time required for 40 g of
this reactant to reduce to 10 g will be [Given that log 10 2 = 0.3010]
Option A 602 s
Option B 230.3 s
Option C 301 s
Option D 2000 s
Correct Option A
Solution:
Half life for first order reaction is calculated by-
1
1
-3
2
1/2 1/2
0.693 0.693
Half life period, t = s
k
2.303 10
= 300.91 s
Now, 40g t 20 g t 10 g
?
?
?
So, 40 g substance requires 2 half-life periods to reduce upto 10 g
Time taken in reduction = 2 × 300.91 s
= 601.82
? 602 s
Q 3. For a reaction, activation energy E a= 0 and the rate constant at 200 K is 1.6 × 10
6
s
–1
. The
rate constant at 400 K will be [Given that gas constant, R = 8.314 J K
–1
mol
–1
]
Option A 3.2 × 10
6
s
–1
Option B 3.2 × 10
4
s
–1
Option C 1.6 × 10
6
s
–1
Option D 1.6 × 10
3
s
–1
Correct Option C
Solution: From Arrhenius equation,
400 a 21
200 1 2
a
400 400
200 200
400 200
kE TT
log
k 2.303R T T
Since, given that E 0
kk
log 0 1
kk
So, k k
?? ?
?
??
??
?
? ? ? ?
?
Page 3
CHEMISTRY
Q 1. Following limiting molar conductivities are given as
?° m (H 2SO 4) = x S cm
2
mol
-1
?° m (K 2SO 4) = y S cm
2
mol
-1
?° m (CH 3COOK) = z S cm
2
mol
-1
?° m (in S cm
2
mol
-1
) for CH 3COOH will be
Option A
(x y)
z
2
?
?
Option B x – y + 2z
Option C x + y + z
Option D x - y + z
Correct Option A
Solution:
Molar conductivity of AB is calculated as-
? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ?
m m m
m 3 m 3 m
m3
m 3 m 2 4 m 2 4
AB A B
So , CH COOH CH COO H
So CH COOH
11
CH COOK H SO K SO
22
xy
z
22
xy
z
2
??
??
? ? ? ? ?
? ? ? ? ?
?
? ? ? ? ? ?
? ? ?
? ??
??
??
??
Q 2. A first order reaction has a rate constant of 2.303 × 10
–3
s
–1
. The time required for 40 g of
this reactant to reduce to 10 g will be [Given that log 10 2 = 0.3010]
Option A 602 s
Option B 230.3 s
Option C 301 s
Option D 2000 s
Correct Option A
Solution:
Half life for first order reaction is calculated by-
1
1
-3
2
1/2 1/2
0.693 0.693
Half life period, t = s
k
2.303 10
= 300.91 s
Now, 40g t 20 g t 10 g
?
?
?
So, 40 g substance requires 2 half-life periods to reduce upto 10 g
Time taken in reduction = 2 × 300.91 s
= 601.82
? 602 s
Q 3. For a reaction, activation energy E a= 0 and the rate constant at 200 K is 1.6 × 10
6
s
–1
. The
rate constant at 400 K will be [Given that gas constant, R = 8.314 J K
–1
mol
–1
]
Option A 3.2 × 10
6
s
–1
Option B 3.2 × 10
4
s
–1
Option C 1.6 × 10
6
s
–1
Option D 1.6 × 10
3
s
–1
Correct Option C
Solution: From Arrhenius equation,
400 a 21
200 1 2
a
400 400
200 200
400 200
kE TT
log
k 2.303R T T
Since, given that E 0
kk
log 0 1
kk
So, k k
?? ?
?
??
??
?
? ? ? ?
?
So rate constant at 400 k = 1.6 × 10
6
s
–1
Q 4. The correct option representing a Freundlich adsorption isotherm is
Option A
1
x
k p
m
?
?
Option B
0.3
x
k p
m
?
Option C
2.5
x
k p
m
?
Option D
0.5
x
k p
m
?
?
Correct Option B
Solution: According to Freundlich isotherm, graph between x/m and P is drawn as-
? ?
1 / n
0.3
x
KP
m
n > 1
x1
So, KP as 0.3
mn
?
??
So, Answer is (B).
Q 5. Which of the following is paramagnetic?
Option A O 2
Option B N 2
Option C H 2
Option D Li 2
Correct Option A
Solution:
Total number of electrons in O 2
8 + 8 = 16 electrons
Distribution of electrons in MO(molecular orbitals) follows the order as
(s1s)
2
, (s*1s)
2
, (s2s)
2
, (s*2s)
2
, (s2p z)
2
,
(p2p x)
2
(p*2p x)
1
(p2p y)
2
' (2p y)
1
So, in O 2 molecule, there are two (2) unpaired electrons, so, it is a "paramagnetic" substance in
nature.
Page 4
CHEMISTRY
Q 1. Following limiting molar conductivities are given as
?° m (H 2SO 4) = x S cm
2
mol
-1
?° m (K 2SO 4) = y S cm
2
mol
-1
?° m (CH 3COOK) = z S cm
2
mol
-1
?° m (in S cm
2
mol
-1
) for CH 3COOH will be
Option A
(x y)
z
2
?
?
Option B x – y + 2z
Option C x + y + z
Option D x - y + z
Correct Option A
Solution:
Molar conductivity of AB is calculated as-
? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ?
m m m
m 3 m 3 m
m3
m 3 m 2 4 m 2 4
AB A B
So , CH COOH CH COO H
So CH COOH
11
CH COOK H SO K SO
22
xy
z
22
xy
z
2
??
??
? ? ? ? ?
? ? ? ? ?
?
? ? ? ? ? ?
? ? ?
? ??
??
??
??
Q 2. A first order reaction has a rate constant of 2.303 × 10
–3
s
–1
. The time required for 40 g of
this reactant to reduce to 10 g will be [Given that log 10 2 = 0.3010]
Option A 602 s
Option B 230.3 s
Option C 301 s
Option D 2000 s
Correct Option A
Solution:
Half life for first order reaction is calculated by-
1
1
-3
2
1/2 1/2
0.693 0.693
Half life period, t = s
k
2.303 10
= 300.91 s
Now, 40g t 20 g t 10 g
?
?
?
So, 40 g substance requires 2 half-life periods to reduce upto 10 g
Time taken in reduction = 2 × 300.91 s
= 601.82
? 602 s
Q 3. For a reaction, activation energy E a= 0 and the rate constant at 200 K is 1.6 × 10
6
s
–1
. The
rate constant at 400 K will be [Given that gas constant, R = 8.314 J K
–1
mol
–1
]
Option A 3.2 × 10
6
s
–1
Option B 3.2 × 10
4
s
–1
Option C 1.6 × 10
6
s
–1
Option D 1.6 × 10
3
s
–1
Correct Option C
Solution: From Arrhenius equation,
400 a 21
200 1 2
a
400 400
200 200
400 200
kE TT
log
k 2.303R T T
Since, given that E 0
kk
log 0 1
kk
So, k k
?? ?
?
??
??
?
? ? ? ?
?
So rate constant at 400 k = 1.6 × 10
6
s
–1
Q 4. The correct option representing a Freundlich adsorption isotherm is
Option A
1
x
k p
m
?
?
Option B
0.3
x
k p
m
?
Option C
2.5
x
k p
m
?
Option D
0.5
x
k p
m
?
?
Correct Option B
Solution: According to Freundlich isotherm, graph between x/m and P is drawn as-
? ?
1 / n
0.3
x
KP
m
n > 1
x1
So, KP as 0.3
mn
?
??
So, Answer is (B).
Q 5. Which of the following is paramagnetic?
Option A O 2
Option B N 2
Option C H 2
Option D Li 2
Correct Option A
Solution:
Total number of electrons in O 2
8 + 8 = 16 electrons
Distribution of electrons in MO(molecular orbitals) follows the order as
(s1s)
2
, (s*1s)
2
, (s2s)
2
, (s*2s)
2
, (s2p z)
2
,
(p2p x)
2
(p*2p x)
1
(p2p y)
2
' (2p y)
1
So, in O 2 molecule, there are two (2) unpaired electrons, so, it is a "paramagnetic" substance in
nature.
Q 6. Which of the following is the correct order of dipole moment?
Option A H 2O < NF 3< NH 3< BF 3
Option B NH 3< BF 3< NF 3< H 2O
Option C BF3< NF 3< NH 3< H 2O
Option D BF 3< NH 3< NF 3< H 2O
Correct Option A
Solution:
Dipole moment of a molecule is the vector sum of dipoles of bonds. So based on molecular geometry
of following molecules,
Three equal vectors at
120° has resultant = 0
so nonpolarmolecule
Vectors not aligned in
the same direction of
lone pair so less dipole
moment
Q 7. Crude sodium chloride obtained by crystallisation of brine solution does not contain
Option A CaSO 4
Option B MgSO 4
Option C Na 2SO 4
Option D MgCl 2
Correct Option B
Solution:
Crude sodium chloride generally obtained by crystallisation of brine solution contains Na 2SO 4, CaSO 4,
CaCl 2 and MgCl 2 as impurities.
Page 5
CHEMISTRY
Q 1. Following limiting molar conductivities are given as
?° m (H 2SO 4) = x S cm
2
mol
-1
?° m (K 2SO 4) = y S cm
2
mol
-1
?° m (CH 3COOK) = z S cm
2
mol
-1
?° m (in S cm
2
mol
-1
) for CH 3COOH will be
Option A
(x y)
z
2
?
?
Option B x – y + 2z
Option C x + y + z
Option D x - y + z
Correct Option A
Solution:
Molar conductivity of AB is calculated as-
? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ?
m m m
m 3 m 3 m
m3
m 3 m 2 4 m 2 4
AB A B
So , CH COOH CH COO H
So CH COOH
11
CH COOK H SO K SO
22
xy
z
22
xy
z
2
??
??
? ? ? ? ?
? ? ? ? ?
?
? ? ? ? ? ?
? ? ?
? ??
??
??
??
Q 2. A first order reaction has a rate constant of 2.303 × 10
–3
s
–1
. The time required for 40 g of
this reactant to reduce to 10 g will be [Given that log 10 2 = 0.3010]
Option A 602 s
Option B 230.3 s
Option C 301 s
Option D 2000 s
Correct Option A
Solution:
Half life for first order reaction is calculated by-
1
1
-3
2
1/2 1/2
0.693 0.693
Half life period, t = s
k
2.303 10
= 300.91 s
Now, 40g t 20 g t 10 g
?
?
?
So, 40 g substance requires 2 half-life periods to reduce upto 10 g
Time taken in reduction = 2 × 300.91 s
= 601.82
? 602 s
Q 3. For a reaction, activation energy E a= 0 and the rate constant at 200 K is 1.6 × 10
6
s
–1
. The
rate constant at 400 K will be [Given that gas constant, R = 8.314 J K
–1
mol
–1
]
Option A 3.2 × 10
6
s
–1
Option B 3.2 × 10
4
s
–1
Option C 1.6 × 10
6
s
–1
Option D 1.6 × 10
3
s
–1
Correct Option C
Solution: From Arrhenius equation,
400 a 21
200 1 2
a
400 400
200 200
400 200
kE TT
log
k 2.303R T T
Since, given that E 0
kk
log 0 1
kk
So, k k
?? ?
?
??
??
?
? ? ? ?
?
So rate constant at 400 k = 1.6 × 10
6
s
–1
Q 4. The correct option representing a Freundlich adsorption isotherm is
Option A
1
x
k p
m
?
?
Option B
0.3
x
k p
m
?
Option C
2.5
x
k p
m
?
Option D
0.5
x
k p
m
?
?
Correct Option B
Solution: According to Freundlich isotherm, graph between x/m and P is drawn as-
? ?
1 / n
0.3
x
KP
m
n > 1
x1
So, KP as 0.3
mn
?
??
So, Answer is (B).
Q 5. Which of the following is paramagnetic?
Option A O 2
Option B N 2
Option C H 2
Option D Li 2
Correct Option A
Solution:
Total number of electrons in O 2
8 + 8 = 16 electrons
Distribution of electrons in MO(molecular orbitals) follows the order as
(s1s)
2
, (s*1s)
2
, (s2s)
2
, (s*2s)
2
, (s2p z)
2
,
(p2p x)
2
(p*2p x)
1
(p2p y)
2
' (2p y)
1
So, in O 2 molecule, there are two (2) unpaired electrons, so, it is a "paramagnetic" substance in
nature.
Q 6. Which of the following is the correct order of dipole moment?
Option A H 2O < NF 3< NH 3< BF 3
Option B NH 3< BF 3< NF 3< H 2O
Option C BF3< NF 3< NH 3< H 2O
Option D BF 3< NH 3< NF 3< H 2O
Correct Option A
Solution:
Dipole moment of a molecule is the vector sum of dipoles of bonds. So based on molecular geometry
of following molecules,
Three equal vectors at
120° has resultant = 0
so nonpolarmolecule
Vectors not aligned in
the same direction of
lone pair so less dipole
moment
Q 7. Crude sodium chloride obtained by crystallisation of brine solution does not contain
Option A CaSO 4
Option B MgSO 4
Option C Na 2SO 4
Option D MgCl 2
Correct Option B
Solution:
Crude sodium chloride generally obtained by crystallisation of brine solution contains Na 2SO 4, CaSO 4,
CaCl 2 and MgCl 2 as impurities.
Q 8. Which of the alkali metal chloride (MCl) forms its dihydrate salt (MCl . 2H 2O) easily?
Option A KCl
Option B LiCl
Option C CsCl
Option D RbCl
Correct Option B
Solution: Out of alkali metal chlorides only LiCl forms a dihydrate, other metal chlorides (KCl, CsCl,
RbCl) do not form hydrates.
Q 9. The reaction that does not give benzoic acid as the major product is
Correct Option D
Solution:
PCC is oxidising agent.
PCC oxidises primary alcohol to aldehyde.
Q 10. The amine that reacts with Hinsberg’s reagent to give an alkali insoluble product is
Option A
Option A
Option B
Option C
Option D
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FAQs on NEET 2019 Past Year Paper with Detailed Solutions - NEET Mock Test Series 2025
| 1. What is NEET exam? |  |
Ans. NEET exam stands for National Eligibility cum Entrance Test. It is a national level entrance exam conducted for admission to undergraduate medical courses (MBBS/BDS) in various government and private medical colleges in India.
| 2. How to prepare for NEET exam? | |
Ans. To prepare for NEET exam, it is important to have a proper study plan, study material, and practice regularly with mock tests. It is also important to revise the concepts regularly and focus on weaker areas. Coaching classes and guidance from experts can also be helpful.
| 3. What is the exam pattern of NEET? | |
Ans. NEET exam is a pen and paper-based exam consisting of 180 multiple-choice questions from Physics, Chemistry, and Biology (Botany and Zoology). The exam is conducted for 720 marks, with each question carrying 4 marks and 1 mark deducted for each wrong answer. The duration of the exam is 3 hours.
| 4. What is the eligibility criteria for NEET exam? | |
Ans. The eligibility criteria for NEET exam include: the candidate should be an Indian citizen or an Overseas Citizen of India (OCI), the candidate should have completed 17 years of age at the time of admission or should complete it on or before December 31 of the year of admission, the candidate should have passed class 12 or equivalent with Physics, Chemistry, Biology/Biotechnology (which shall include practical tests in these subjects) and English.
| 5. What is the cutoff for NEET exam? | |
Ans. The cutoff for NEET exam varies every year and depends on factors such as the difficulty level of the exam, number of candidates appearing for the exam, and availability of seats. The cutoff is the minimum marks required to qualify for the exam and is usually around 50th percentile. However, it is important to aim for a higher score to secure admission to a good medical college.
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