JEE Main 2020 Chemistry September 2 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The increasing order of the following compounds towards HCN addition is:
H CO
3
CHO
(i)
NO
2
CHO
(ii)
OCH
3
CHO
(iii)
O N
2
CHO
(iv)
(1) (iii) < (i) < (iv) < (ii) (2) (iii) < (iv) < (i) < (ii)
(3) (i) < (iii) < (iv) < (ii) (4) (iii) < (iv) < (ii) < (i)
Sol. 1
In HCN, CN
–
 is acts as nucleophile, attack first that –CHO group which has maximum
positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So
reactivity order will be
CHO
NO
(–M)
2
NO
(–I)
2
>
CHO
 >
OCH
(–I)
3
OCH
(+M)
3
CHO
>
CHO
So, option (1) is correct answer.
2. Which of the following is used for the preparation of colloids?
(1) Van Arkel Method (2) Ostwald Process
(3) Mond Process (4) Bredig’s Arc Method
Sol. 4
Bredig’s Arc method
Chapter name surface chemistry
3. An open beaker of water in equilibrium with water vapour is in a sealed container.
When a few grams of glucose are added to the beaker of water, the rate at which water
molecules:
(1) leaves the vapour increases (2) leaves the solution increases
(3) leaves the vapour decreases (4) leaves the solution decreases
Sol. 1
×
×
×
×
×
×
×
×
×
×
×
×
× ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Vap. press = Pº
× ×
×
×
×
×
×
×
×
Vap. press = P
s
×××
×
H
2
O (l) 
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
H
2
O(g) H
2
O( ) ? ? ? ? ? ?
? ? ? ? ?
H
2
O(g)
K
p
 = Pº K
P
 = P
s
Backward shift
    vapours ?
Hence Rate at which water molecules leaves the vap. increases.
Page 2


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The increasing order of the following compounds towards HCN addition is:
H CO
3
CHO
(i)
NO
2
CHO
(ii)
OCH
3
CHO
(iii)
O N
2
CHO
(iv)
(1) (iii) < (i) < (iv) < (ii) (2) (iii) < (iv) < (i) < (ii)
(3) (i) < (iii) < (iv) < (ii) (4) (iii) < (iv) < (ii) < (i)
Sol. 1
In HCN, CN
–
 is acts as nucleophile, attack first that –CHO group which has maximum
positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So
reactivity order will be
CHO
NO
(–M)
2
NO
(–I)
2
>
CHO
 >
OCH
(–I)
3
OCH
(+M)
3
CHO
>
CHO
So, option (1) is correct answer.
2. Which of the following is used for the preparation of colloids?
(1) Van Arkel Method (2) Ostwald Process
(3) Mond Process (4) Bredig’s Arc Method
Sol. 4
Bredig’s Arc method
Chapter name surface chemistry
3. An open beaker of water in equilibrium with water vapour is in a sealed container.
When a few grams of glucose are added to the beaker of water, the rate at which water
molecules:
(1) leaves the vapour increases (2) leaves the solution increases
(3) leaves the vapour decreases (4) leaves the solution decreases
Sol. 1
×
×
×
×
×
×
×
×
×
×
×
×
× ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Vap. press = Pº
× ×
×
×
×
×
×
×
×
Vap. press = P
s
×××
×
H
2
O (l) 
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
H
2
O(g) H
2
O( ) ? ? ? ? ? ?
? ? ? ? ?
H
2
O(g)
K
p
 = Pº K
P
 = P
s
Backward shift
    vapours ?
Hence Rate at which water molecules leaves the vap. increases.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 2
4. For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following state-
ments:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are:
(1) (I), (III) and (IV) only (2) (I), (II) and (III) only
(3) (II), (III) and (IV) only (4) (I) and (II) only
Sol. 2
Mn
2+
 [Ar]3d
5
 it can form low spin as well as high spin complex depending upon nature
of ligand same of Ni
2+
 ion with coordination no 4. It can be dsp
2
 or sp
3
 i:e low spin or
high spin depending open nature of ligand.
5. The statement that is not true about ozone is:
(1) in the stratosphere, it forms a protective shield against UV radiation.
(2) in the atmosphere, it is depleted by CFCs.
(3) in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O
3
 to
give chlorine dioxide radicals.
(4) it is a toxic gas and its reaction with NO gives NO
2
.
Sol. 3
Cl
?
 + O
3
 
? ? ? ?
Cl O
?
 + O
2
Chlorine monoxide
Hence option (3)
6. Consider the following reactions:
(i) Glucose + ROH ? ? ? ? ? ?
HCl dry
 Acetal 
O ) CO CH (
of eq. x
2 3
? ? ? ? ? ? acetyl derivative e
(ii) Glucose ? ? ? ? ?
2
H / Ni
 A 
O ) CO (CH
of eq. y
2 3
? ? ? ? ? ? acetyl derivative e
(iii)Glucose 
O ) CO (CH
of eq. z
2 3
? ? ? ? ? ? acetyl derivative e
'x', 'y' and 'z' in these reactions are respectively.
(1) 4, 5 & 5 (2) 5, 4 & 5
(3) 5, 6 & 5 (4) 4, 6 & 5
Page 3


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The increasing order of the following compounds towards HCN addition is:
H CO
3
CHO
(i)
NO
2
CHO
(ii)
OCH
3
CHO
(iii)
O N
2
CHO
(iv)
(1) (iii) < (i) < (iv) < (ii) (2) (iii) < (iv) < (i) < (ii)
(3) (i) < (iii) < (iv) < (ii) (4) (iii) < (iv) < (ii) < (i)
Sol. 1
In HCN, CN
–
 is acts as nucleophile, attack first that –CHO group which has maximum
positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So
reactivity order will be
CHO
NO
(–M)
2
NO
(–I)
2
>
CHO
 >
OCH
(–I)
3
OCH
(+M)
3
CHO
>
CHO
So, option (1) is correct answer.
2. Which of the following is used for the preparation of colloids?
(1) Van Arkel Method (2) Ostwald Process
(3) Mond Process (4) Bredig’s Arc Method
Sol. 4
Bredig’s Arc method
Chapter name surface chemistry
3. An open beaker of water in equilibrium with water vapour is in a sealed container.
When a few grams of glucose are added to the beaker of water, the rate at which water
molecules:
(1) leaves the vapour increases (2) leaves the solution increases
(3) leaves the vapour decreases (4) leaves the solution decreases
Sol. 1
×
×
×
×
×
×
×
×
×
×
×
×
× ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Vap. press = Pº
× ×
×
×
×
×
×
×
×
Vap. press = P
s
×××
×
H
2
O (l) 
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
H
2
O(g) H
2
O( ) ? ? ? ? ? ?
? ? ? ? ?
H
2
O(g)
K
p
 = Pº K
P
 = P
s
Backward shift
    vapours ?
Hence Rate at which water molecules leaves the vap. increases.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 2
4. For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following state-
ments:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are:
(1) (I), (III) and (IV) only (2) (I), (II) and (III) only
(3) (II), (III) and (IV) only (4) (I) and (II) only
Sol. 2
Mn
2+
 [Ar]3d
5
 it can form low spin as well as high spin complex depending upon nature
of ligand same of Ni
2+
 ion with coordination no 4. It can be dsp
2
 or sp
3
 i:e low spin or
high spin depending open nature of ligand.
5. The statement that is not true about ozone is:
(1) in the stratosphere, it forms a protective shield against UV radiation.
(2) in the atmosphere, it is depleted by CFCs.
(3) in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O
3
 to
give chlorine dioxide radicals.
(4) it is a toxic gas and its reaction with NO gives NO
2
.
Sol. 3
Cl
?
 + O
3
 
? ? ? ?
Cl O
?
 + O
2
Chlorine monoxide
Hence option (3)
6. Consider the following reactions:
(i) Glucose + ROH ? ? ? ? ? ?
HCl dry
 Acetal 
O ) CO CH (
of eq. x
2 3
? ? ? ? ? ? acetyl derivative e
(ii) Glucose ? ? ? ? ?
2
H / Ni
 A 
O ) CO (CH
of eq. y
2 3
? ? ? ? ? ? acetyl derivative e
(iii)Glucose 
O ) CO (CH
of eq. z
2 3
? ? ? ? ? ? acetyl derivative e
'x', 'y' and 'z' in these reactions are respectively.
(1) 4, 5 & 5 (2) 5, 4 & 5
(3) 5, 6 & 5 (4) 4, 6 & 5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) Glucose + ROH 
dry HCl
? ? ? ? ?
H—C OH —
HO—C H —
H—C OH —
H—C
C
H OR
CH OH
2
O
 
? ? 3 2
4 eqof
CH CO O
acetyl derivative ? ? ? ? ? ?
(ii) Glucose 
2
Ni/H
? ? ? ? ?
CH OH
2
(CHOH)
4
CH OH
2
? ? 3 2
6 eq of
CH CO O
? ? ? ? ? ?
 acetyl derivative e
(iii) Glucose 
? ?
3 2
5 eq. of
CH CO O
? ? ? ? ? ?
 Acetyl derivative e
(CH
3
CO)
2
O reacts with –OH group to form acetyl derivative, so as the no. of –OH group
no. of eq. of (CH
3
CO)
2
O will be used
So, x = 4
y = 6
z = 5
So, option (4) will be correct answer.
7. The IUPAC name for the following compound is:
CHO
H C
3
COOH
CH
3
(1) 2,5-dimethyl-5-carboxy-hex-3-enal (2) 2,5-dimethyl-6-oxo-hex-3-enoic acid
(3) 6-formyl-2-methyl-hex-3-enoic acid (4) 2,5-dimethyl-6-carboxy-hex-3-enal
Sol. 2
H C
3
CHO
COOH
CH
3
6
5
4
3
2
1
2,5–Dimethyl–6–oxohex–3–enoic acid
Page 4


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The increasing order of the following compounds towards HCN addition is:
H CO
3
CHO
(i)
NO
2
CHO
(ii)
OCH
3
CHO
(iii)
O N
2
CHO
(iv)
(1) (iii) < (i) < (iv) < (ii) (2) (iii) < (iv) < (i) < (ii)
(3) (i) < (iii) < (iv) < (ii) (4) (iii) < (iv) < (ii) < (i)
Sol. 1
In HCN, CN
–
 is acts as nucleophile, attack first that –CHO group which has maximum
positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So
reactivity order will be
CHO
NO
(–M)
2
NO
(–I)
2
>
CHO
 >
OCH
(–I)
3
OCH
(+M)
3
CHO
>
CHO
So, option (1) is correct answer.
2. Which of the following is used for the preparation of colloids?
(1) Van Arkel Method (2) Ostwald Process
(3) Mond Process (4) Bredig’s Arc Method
Sol. 4
Bredig’s Arc method
Chapter name surface chemistry
3. An open beaker of water in equilibrium with water vapour is in a sealed container.
When a few grams of glucose are added to the beaker of water, the rate at which water
molecules:
(1) leaves the vapour increases (2) leaves the solution increases
(3) leaves the vapour decreases (4) leaves the solution decreases
Sol. 1
×
×
×
×
×
×
×
×
×
×
×
×
× ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Vap. press = Pº
× ×
×
×
×
×
×
×
×
Vap. press = P
s
×××
×
H
2
O (l) 
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
H
2
O(g) H
2
O( ) ? ? ? ? ? ?
? ? ? ? ?
H
2
O(g)
K
p
 = Pº K
P
 = P
s
Backward shift
    vapours ?
Hence Rate at which water molecules leaves the vap. increases.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 2
4. For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following state-
ments:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are:
(1) (I), (III) and (IV) only (2) (I), (II) and (III) only
(3) (II), (III) and (IV) only (4) (I) and (II) only
Sol. 2
Mn
2+
 [Ar]3d
5
 it can form low spin as well as high spin complex depending upon nature
of ligand same of Ni
2+
 ion with coordination no 4. It can be dsp
2
 or sp
3
 i:e low spin or
high spin depending open nature of ligand.
5. The statement that is not true about ozone is:
(1) in the stratosphere, it forms a protective shield against UV radiation.
(2) in the atmosphere, it is depleted by CFCs.
(3) in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O
3
 to
give chlorine dioxide radicals.
(4) it is a toxic gas and its reaction with NO gives NO
2
.
Sol. 3
Cl
?
 + O
3
 
? ? ? ?
Cl O
?
 + O
2
Chlorine monoxide
Hence option (3)
6. Consider the following reactions:
(i) Glucose + ROH ? ? ? ? ? ?
HCl dry
 Acetal 
O ) CO CH (
of eq. x
2 3
? ? ? ? ? ? acetyl derivative e
(ii) Glucose ? ? ? ? ?
2
H / Ni
 A 
O ) CO (CH
of eq. y
2 3
? ? ? ? ? ? acetyl derivative e
(iii)Glucose 
O ) CO (CH
of eq. z
2 3
? ? ? ? ? ? acetyl derivative e
'x', 'y' and 'z' in these reactions are respectively.
(1) 4, 5 & 5 (2) 5, 4 & 5
(3) 5, 6 & 5 (4) 4, 6 & 5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) Glucose + ROH 
dry HCl
? ? ? ? ?
H—C OH —
HO—C H —
H—C OH —
H—C
C
H OR
CH OH
2
O
 
? ? 3 2
4 eqof
CH CO O
acetyl derivative ? ? ? ? ? ?
(ii) Glucose 
2
Ni/H
? ? ? ? ?
CH OH
2
(CHOH)
4
CH OH
2
? ? 3 2
6 eq of
CH CO O
? ? ? ? ? ?
 acetyl derivative e
(iii) Glucose 
? ?
3 2
5 eq. of
CH CO O
? ? ? ? ? ?
 Acetyl derivative e
(CH
3
CO)
2
O reacts with –OH group to form acetyl derivative, so as the no. of –OH group
no. of eq. of (CH
3
CO)
2
O will be used
So, x = 4
y = 6
z = 5
So, option (4) will be correct answer.
7. The IUPAC name for the following compound is:
CHO
H C
3
COOH
CH
3
(1) 2,5-dimethyl-5-carboxy-hex-3-enal (2) 2,5-dimethyl-6-oxo-hex-3-enoic acid
(3) 6-formyl-2-methyl-hex-3-enoic acid (4) 2,5-dimethyl-6-carboxy-hex-3-enal
Sol. 2
H C
3
CHO
COOH
CH
3
6
5
4
3
2
1
2,5–Dimethyl–6–oxohex–3–enoic acid
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 4
8. For the following Assertion and Reason, the correct option is
Assertion (A): When Cu (II) and sulphide ions are mixed, they react together ex-
tremely quickly to give a solid.
Reason (R): The equilibrium constant of Cu
2+ 
(aq) + S
2–
 (aq)  CuS (s) is high
because the solubility product is low.
(1) (A) is false and (R) is true.
(2) Both (A) and (R) are false.
(3) Both (A) and (R) are true but (R) is not the explanation for (A).
(4) Both (A) and (R) are true but (R) is the explanation for (A).
Sol. 4
(A) is (B) true &
(R) is correct explanation of (A)
Ans. 4
9. Which one of the following graphs is not correct for ideal gas?
d
T
I
d
T
II
d
1/T
III
d
P
IV
d = Density, P = Pressure, T = Temperature
(1) I (2) IV (3) III (4) II
Sol. 4
For ideal Gas
d = 
P M
RT
?
d v/s T ? Hyperbolic
T
d
d v/s 
1
T
? ?St. line
1/T
d
d v/s p ? ?St line
P
d
? ‘II’ Graph is incorrect
Ans (4)
Page 5


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 1
Date : 2
nd
 September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Chemistry
1. The increasing order of the following compounds towards HCN addition is:
H CO
3
CHO
(i)
NO
2
CHO
(ii)
OCH
3
CHO
(iii)
O N
2
CHO
(iv)
(1) (iii) < (i) < (iv) < (ii) (2) (iii) < (iv) < (i) < (ii)
(3) (i) < (iii) < (iv) < (ii) (4) (iii) < (iv) < (ii) < (i)
Sol. 1
In HCN, CN
–
 is acts as nucleophile, attack first that –CHO group which has maximum
positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So
reactivity order will be
CHO
NO
(–M)
2
NO
(–I)
2
>
CHO
 >
OCH
(–I)
3
OCH
(+M)
3
CHO
>
CHO
So, option (1) is correct answer.
2. Which of the following is used for the preparation of colloids?
(1) Van Arkel Method (2) Ostwald Process
(3) Mond Process (4) Bredig’s Arc Method
Sol. 4
Bredig’s Arc method
Chapter name surface chemistry
3. An open beaker of water in equilibrium with water vapour is in a sealed container.
When a few grams of glucose are added to the beaker of water, the rate at which water
molecules:
(1) leaves the vapour increases (2) leaves the solution increases
(3) leaves the vapour decreases (4) leaves the solution decreases
Sol. 1
×
×
×
×
×
×
×
×
×
×
×
×
× ×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
Vap. press = Pº
× ×
×
×
×
×
×
×
×
Vap. press = P
s
×××
×
H
2
O (l) 
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
H
2
O(g) H
2
O( ) ? ? ? ? ? ?
? ? ? ? ?
H
2
O(g)
K
p
 = Pº K
P
 = P
s
Backward shift
    vapours ?
Hence Rate at which water molecules leaves the vap. increases.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 2
4. For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following state-
ments:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are:
(1) (I), (III) and (IV) only (2) (I), (II) and (III) only
(3) (II), (III) and (IV) only (4) (I) and (II) only
Sol. 2
Mn
2+
 [Ar]3d
5
 it can form low spin as well as high spin complex depending upon nature
of ligand same of Ni
2+
 ion with coordination no 4. It can be dsp
2
 or sp
3
 i:e low spin or
high spin depending open nature of ligand.
5. The statement that is not true about ozone is:
(1) in the stratosphere, it forms a protective shield against UV radiation.
(2) in the atmosphere, it is depleted by CFCs.
(3) in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O
3
 to
give chlorine dioxide radicals.
(4) it is a toxic gas and its reaction with NO gives NO
2
.
Sol. 3
Cl
?
 + O
3
 
? ? ? ?
Cl O
?
 + O
2
Chlorine monoxide
Hence option (3)
6. Consider the following reactions:
(i) Glucose + ROH ? ? ? ? ? ?
HCl dry
 Acetal 
O ) CO CH (
of eq. x
2 3
? ? ? ? ? ? acetyl derivative e
(ii) Glucose ? ? ? ? ?
2
H / Ni
 A 
O ) CO (CH
of eq. y
2 3
? ? ? ? ? ? acetyl derivative e
(iii)Glucose 
O ) CO (CH
of eq. z
2 3
? ? ? ? ? ? acetyl derivative e
'x', 'y' and 'z' in these reactions are respectively.
(1) 4, 5 & 5 (2) 5, 4 & 5
(3) 5, 6 & 5 (4) 4, 6 & 5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 3
Sol. 4
(i) Glucose + ROH 
dry HCl
? ? ? ? ?
H—C OH —
HO—C H —
H—C OH —
H—C
C
H OR
CH OH
2
O
 
? ? 3 2
4 eqof
CH CO O
acetyl derivative ? ? ? ? ? ?
(ii) Glucose 
2
Ni/H
? ? ? ? ?
CH OH
2
(CHOH)
4
CH OH
2
? ? 3 2
6 eq of
CH CO O
? ? ? ? ? ?
 acetyl derivative e
(iii) Glucose 
? ?
3 2
5 eq. of
CH CO O
? ? ? ? ? ?
 Acetyl derivative e
(CH
3
CO)
2
O reacts with –OH group to form acetyl derivative, so as the no. of –OH group
no. of eq. of (CH
3
CO)
2
O will be used
So, x = 4
y = 6
z = 5
So, option (4) will be correct answer.
7. The IUPAC name for the following compound is:
CHO
H C
3
COOH
CH
3
(1) 2,5-dimethyl-5-carboxy-hex-3-enal (2) 2,5-dimethyl-6-oxo-hex-3-enoic acid
(3) 6-formyl-2-methyl-hex-3-enoic acid (4) 2,5-dimethyl-6-carboxy-hex-3-enal
Sol. 2
H C
3
CHO
COOH
CH
3
6
5
4
3
2
1
2,5–Dimethyl–6–oxohex–3–enoic acid
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 4
8. For the following Assertion and Reason, the correct option is
Assertion (A): When Cu (II) and sulphide ions are mixed, they react together ex-
tremely quickly to give a solid.
Reason (R): The equilibrium constant of Cu
2+ 
(aq) + S
2–
 (aq)  CuS (s) is high
because the solubility product is low.
(1) (A) is false and (R) is true.
(2) Both (A) and (R) are false.
(3) Both (A) and (R) are true but (R) is not the explanation for (A).
(4) Both (A) and (R) are true but (R) is the explanation for (A).
Sol. 4
(A) is (B) true &
(R) is correct explanation of (A)
Ans. 4
9. Which one of the following graphs is not correct for ideal gas?
d
T
I
d
T
II
d
1/T
III
d
P
IV
d = Density, P = Pressure, T = Temperature
(1) I (2) IV (3) III (4) II
Sol. 4
For ideal Gas
d = 
P M
RT
?
d v/s T ? Hyperbolic
T
d
d v/s 
1
T
? ?St. line
1/T
d
d v/s p ? ?St line
P
d
? ‘II’ Graph is incorrect
Ans (4)
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-1), Chemistry     Page | 5
10. While titrating dilute HCl solution with aqueous NaOH, which of the following will not be
required?
(1) Bunsen burner and measuring cylinder (2) Burette and porcelain tile
(3) Clamp and phenolphthalein (4) Pipette and distilled water
Sol. 1
Bunsen Burner & measuring cylinder are not Required. As titration is already on
exothermic process
Ans.(1)
11. In Carius method of estimation of halogen, 0.172 g of an organic compound showed
presence of 0.08 g of bromine. Which of these is the correct structure of the com-
pound?
(1) H
3
C–Br (2) 
NH
2
Br
Br
(3) 
NH
2
Br
(4) H
3
C–CH
2
–Br
Sol. 3
carius method
mass % of ‘Br’ = 
0.08 8000
100 46.51%
0.172 172
? ? ?
option (1) mass % = 
80
100
95
?
(2) mass % = 
2 80 100
252
? ?
(3) mass % = 
1 80 100
80 72 6 14
? ?
? ? ?
=
8000
%
172
(4) mass % = 
1 80 100
%
109
? ?
Option (3) matches with the given mass percentage value
Ans (3)
12. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when
treated with H
2
 in the presence of a catalyst gives another gas (C) which is basic in
nature. (A) should not be:
(1) (NH
4
)
2
Cr
2
O
7
(2) NaN
3
(3) NH
4
NO
2
(4) Pb(NO
3
)
2
Sol. 4
The gas (B) is N
2
 which is found in air
N
2
 + 3H
2
 
Fe/Mo
 2NH
3
 (Haber's process)
(Basic in nature)
NH
3
 + H
2
O ? NH
4
OH (weak base)
(NH
4
)
2
Cr
2
O
7
 ?? N
2
 + Cr
2
O
3
 + H
2
O
NaN
3
 ?? N
2
 + Na
NH
4
NO
2
 ?? N
2
 + H
2
O
Pb(NO
3
)
2
 ?? PbO + NO
2
 + O
2