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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