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NEET Mock Test Series & Past Year Papers

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 Page 1


1
1. Under isothermal condition, a gas at 300 K expands from 0.1L to 0.25L against a constant external pressure
of 2 bar. The work done by the gas is :-
[Given that 1L bar = 100 J]
(1) –30 J (2) 5kJ (3) 25 J (4) 30 J
Ans. (1)
Sol. W = –P
ext
 (V
2
–V
1
)
P
ext 
 = 2 bar
V
1
 = 0.1 L
V
2
 = 0.25 L
W = –2 bar[0.25 – 0.1] L
W = –2 × 0.15 bar L
W = –0.30 bar L
W = (–0.30) × 100 = –30 J
2. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and
the cations occupy 75% of octahedral voids. The formula of the compound is :-
(1) C
2
A
3
(2) C
3
A
2
(3) C
3
A
4
(4) C
4
A
3
Ans. (3)
Sol. Anion A in HCP
No of ions of A in Unit cell = 6
No of Octahedral voids = 6
75% is occupied by cations C
No of cations C = ?
75
6
100
= ?
3
6
4
= 
9
2
C
9/2
A
6
C
9
A
12
Simple ratio C
3
A
4
3. 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
Ans. (1)
Sol. Ca(OH)
2
(s) ??Ca
+2
(aq) + 2OH
–
(aq)
   S            2S
pH = 9 ;  pOH = 5 ; [OH
–
] = 10
–5
 = 2S
?
?
5
10
S
2
K
sp
 = [Ca
+2
] [OH
–
]
2
K
sp
 = S × (2S)
2
K
sp
 = 4S
3
K
sp
 = 
?
? ?
?
? ?
? ?
3
5
10
4
2
K
sp
= 0.5 × 10
–15
FINAL NEET(UG)–2019 EXAMINATION
(Held On Sunday 05
th
 MAY, 2019)
CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION
Page 2


1
1. Under isothermal condition, a gas at 300 K expands from 0.1L to 0.25L against a constant external pressure
of 2 bar. The work done by the gas is :-
[Given that 1L bar = 100 J]
(1) –30 J (2) 5kJ (3) 25 J (4) 30 J
Ans. (1)
Sol. W = –P
ext
 (V
2
–V
1
)
P
ext 
 = 2 bar
V
1
 = 0.1 L
V
2
 = 0.25 L
W = –2 bar[0.25 – 0.1] L
W = –2 × 0.15 bar L
W = –0.30 bar L
W = (–0.30) × 100 = –30 J
2. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and
the cations occupy 75% of octahedral voids. The formula of the compound is :-
(1) C
2
A
3
(2) C
3
A
2
(3) C
3
A
4
(4) C
4
A
3
Ans. (3)
Sol. Anion A in HCP
No of ions of A in Unit cell = 6
No of Octahedral voids = 6
75% is occupied by cations C
No of cations C = ?
75
6
100
= ?
3
6
4
= 
9
2
C
9/2
A
6
C
9
A
12
Simple ratio C
3
A
4
3. 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
Ans. (1)
Sol. Ca(OH)
2
(s) ??Ca
+2
(aq) + 2OH
–
(aq)
   S            2S
pH = 9 ;  pOH = 5 ; [OH
–
] = 10
–5
 = 2S
?
?
5
10
S
2
K
sp
 = [Ca
+2
] [OH
–
]
2
K
sp
 = S × (2S)
2
K
sp
 = 4S
3
K
sp
 = 
?
? ?
?
? ?
? ?
3
5
10
4
2
K
sp
= 0.5 × 10
–15
FINAL NEET(UG)–2019 EXAMINATION
(Held On Sunday 05
th
 MAY, 2019)
CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION
2
 
4. 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
Ans. (3)
Sol. N
2
(g) + 3H
2
(g) ??2NH
3
(g)
2 mole NH
3
(g) requires 3mole H
2
(g)
20 mole NH
3
(g) requires
= 
?
2
3
20 mole H (g)
2
= 30 mole
5. For an ideal solution, the correct option is :-
(1) ?
mix
 S = 0 at constant T and P (2) ?
mix
 V ? 0 at constant T and P
(3) ?
mix
 H = 0 at constant T and P (4) ?
mix
 G = 0 at constant T and P
Ans. (3)
Sol. For ideal solution ?H
mix
 = 0
6. For a cell involving one electron E
?
cell
 = 0.59V at 298 K, the equilibrium constant for the cell reaction is :-
? ?
? ?
? ?
? ?
2.303RT
Given that 0.059V at T 298K
F
(1) 1.0 × 10
2
(2) 1.0 × 10
5
(3) 1.0 × 10
10
(4) 1.0 × 10
30
Ans. (3)
Sol. E
cell
 = E
o
cell
 – 
10
2.303 RT
log Q
nF
at equlibrium E
cell
 = 0, Q = K
eq.
? ?
o
cell 10 eq.
0.0591
0 E log K
1
E
o
cell
 = +0.0591 log
10
 K
eq.
0.59 = + 0.0591 log
10
 K
eq.
+10 = log
10
 K
eq.
K
eq.
 = 10
+10
7. Among the following, the one that is not a green house gas is :-
(1) nitrous oxide (2) methane (3) ozone (4) sulphur dioxide
Ans. (4)
Sol. Besides carbon dioxide, other greenhouse gases are methane, water vapour, nitrous oxide, CFCs and ozone.
8. The number of sigma ( ?) and pi ( ?) bonds in pent-2-en-4-yne is :-
(1) 10 ? bonds and 3 ? bonds (2) 8 ? bonds and 5 ? bonds
(3) 11 ? bonds and 2 ? bonds (4) 13 ? bonds and no ? bond
Ans. (1)
Sol. H–C–C=C–C C–H ?
H
H H H
Number of sigma bonds = 10
Number of ?-bonds = 3
Page 3


1
1. Under isothermal condition, a gas at 300 K expands from 0.1L to 0.25L against a constant external pressure
of 2 bar. The work done by the gas is :-
[Given that 1L bar = 100 J]
(1) –30 J (2) 5kJ (3) 25 J (4) 30 J
Ans. (1)
Sol. W = –P
ext
 (V
2
–V
1
)
P
ext 
 = 2 bar
V
1
 = 0.1 L
V
2
 = 0.25 L
W = –2 bar[0.25 – 0.1] L
W = –2 × 0.15 bar L
W = –0.30 bar L
W = (–0.30) × 100 = –30 J
2. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and
the cations occupy 75% of octahedral voids. The formula of the compound is :-
(1) C
2
A
3
(2) C
3
A
2
(3) C
3
A
4
(4) C
4
A
3
Ans. (3)
Sol. Anion A in HCP
No of ions of A in Unit cell = 6
No of Octahedral voids = 6
75% is occupied by cations C
No of cations C = ?
75
6
100
= ?
3
6
4
= 
9
2
C
9/2
A
6
C
9
A
12
Simple ratio C
3
A
4
3. 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
Ans. (1)
Sol. Ca(OH)
2
(s) ??Ca
+2
(aq) + 2OH
–
(aq)
   S            2S
pH = 9 ;  pOH = 5 ; [OH
–
] = 10
–5
 = 2S
?
?
5
10
S
2
K
sp
 = [Ca
+2
] [OH
–
]
2
K
sp
 = S × (2S)
2
K
sp
 = 4S
3
K
sp
 = 
?
? ?
?
? ?
? ?
3
5
10
4
2
K
sp
= 0.5 × 10
–15
FINAL NEET(UG)–2019 EXAMINATION
(Held On Sunday 05
th
 MAY, 2019)
CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION
2
 
4. 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
Ans. (3)
Sol. N
2
(g) + 3H
2
(g) ??2NH
3
(g)
2 mole NH
3
(g) requires 3mole H
2
(g)
20 mole NH
3
(g) requires
= 
?
2
3
20 mole H (g)
2
= 30 mole
5. For an ideal solution, the correct option is :-
(1) ?
mix
 S = 0 at constant T and P (2) ?
mix
 V ? 0 at constant T and P
(3) ?
mix
 H = 0 at constant T and P (4) ?
mix
 G = 0 at constant T and P
Ans. (3)
Sol. For ideal solution ?H
mix
 = 0
6. For a cell involving one electron E
?
cell
 = 0.59V at 298 K, the equilibrium constant for the cell reaction is :-
? ?
? ?
? ?
? ?
2.303RT
Given that 0.059V at T 298K
F
(1) 1.0 × 10
2
(2) 1.0 × 10
5
(3) 1.0 × 10
10
(4) 1.0 × 10
30
Ans. (3)
Sol. E
cell
 = E
o
cell
 – 
10
2.303 RT
log Q
nF
at equlibrium E
cell
 = 0, Q = K
eq.
? ?
o
cell 10 eq.
0.0591
0 E log K
1
E
o
cell
 = +0.0591 log
10
 K
eq.
0.59 = + 0.0591 log
10
 K
eq.
+10 = log
10
 K
eq.
K
eq.
 = 10
+10
7. Among the following, the one that is not a green house gas is :-
(1) nitrous oxide (2) methane (3) ozone (4) sulphur dioxide
Ans. (4)
Sol. Besides carbon dioxide, other greenhouse gases are methane, water vapour, nitrous oxide, CFCs and ozone.
8. The number of sigma ( ?) and pi ( ?) bonds in pent-2-en-4-yne is :-
(1) 10 ? bonds and 3 ? bonds (2) 8 ? bonds and 5 ? bonds
(3) 11 ? bonds and 2 ? bonds (4) 13 ? bonds and no ? bond
Ans. (1)
Sol. H–C–C=C–C C–H ?
H
H H H
Number of sigma bonds = 10
Number of ?-bonds = 3
3
9. Which of the following diatomic molecular species has only ? bonds according to Molecular Orbital Theory ?
(1) O
2
(2) N
2
(3) C
2
(4) Be
2
Ans. (3)
Sol. According to M.O.T. electronic configuration of C
2
 molecule is -
?1s
2
 < ?*1s
2
 < ?2s
2
 < ?*2s
2
 < ?
2
p
x
2
= ?
2 
p
y
2
so, C
2
 molecule contain only ' ?' bond
10. Which of the following reactions are disproportionation reaction ?
(a) 2Cu
+
 ??Cu
2+
 + Cu
0
(b) 3MnO
4
2–
 + 4H
+
 ??2MnO
4
–
 + MnO
2
 + 2H
2
O
(c) 2KMnO
4
?
?? ?
 K
2
MnO
4
 + MnO
2
 + O
2
(d) 2MnO
4
–
 + 3Mn
2+
 + 2H
2
O ??5MnO
2
 + 4H
?
Select the correct option from the following :-
(1) (a) and (b) only (2) (a), (b) and (c) (3) (a), (c) and (d) (4) (a) and (d) only
Ans. (1)
Sol. (a) 2Cu
+
 ??Cu
+2
 + Cu
? ?
?
? ? ?
? ?
?
? ?
2
Cu Cu (oxidation)
Cu Cu (Reduction)
(b) MnO
4
2–
 ??MnO
4
–
 (oxidation)
+6          +7
MnO
4
2–
 ? ?MnO
2
 (Reduction)
+6            +4
The above two reaction are disproportionation.
11. Among the following, the narrow spectrum antibiotic is :-
(1) penicillin G (2) ampicillin (3) amoxycillin (4) chloramphenicol
Ans. (1)
Sol. The antibiotics which effective mainly against Gram-positive or Gram-negative bacteria are narrow spectrum
antibiotics. Penicillin G has a narrow spectrum.
ampicillin, amoxycillin, chloramphenicol are broad spectrum antibiotics.
12. The correct order of the basic strength of methyl substituted amines in aqueous solution is :-
(1) (CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N (2) (CH
3
)
3
N>CH
3
NH
2
 > (CH
3
)
2
NH
(3) (CH
3
)
3
N>(CH
3
)
2
NH>CH
3
NH
2
(4) CH
3
NH
2
>(CH
3
)
2
NH > (CH
3
)
3
N
Ans. (1)
Sol. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution
is as follows :
(C
2
H
5
)
2
 NH > (C
2
H
5
)
3
N > C
2
H
5
NH
2
 >NH
3
(CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N > NH
3
13. Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI] I
–
 sol. ?
(1) 50 mL of 1M AgNO
3
 + 50 mL of 1.5 M KI
(2) 50 mL of 1M 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
Ans. (1,2)
Sol. In negatively charged colloid [AgI] I
–
, I
–
 is preferentially adsorbed.
AgNO
3
 + KI ??AgI + KNO
3
When KI is in excess, I
–
 will be adsorbed on the surface of AgI and [AgI] I
–
 is formed
Page 4


1
1. Under isothermal condition, a gas at 300 K expands from 0.1L to 0.25L against a constant external pressure
of 2 bar. The work done by the gas is :-
[Given that 1L bar = 100 J]
(1) –30 J (2) 5kJ (3) 25 J (4) 30 J
Ans. (1)
Sol. W = –P
ext
 (V
2
–V
1
)
P
ext 
 = 2 bar
V
1
 = 0.1 L
V
2
 = 0.25 L
W = –2 bar[0.25 – 0.1] L
W = –2 × 0.15 bar L
W = –0.30 bar L
W = (–0.30) × 100 = –30 J
2. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and
the cations occupy 75% of octahedral voids. The formula of the compound is :-
(1) C
2
A
3
(2) C
3
A
2
(3) C
3
A
4
(4) C
4
A
3
Ans. (3)
Sol. Anion A in HCP
No of ions of A in Unit cell = 6
No of Octahedral voids = 6
75% is occupied by cations C
No of cations C = ?
75
6
100
= ?
3
6
4
= 
9
2
C
9/2
A
6
C
9
A
12
Simple ratio C
3
A
4
3. 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
Ans. (1)
Sol. Ca(OH)
2
(s) ??Ca
+2
(aq) + 2OH
–
(aq)
   S            2S
pH = 9 ;  pOH = 5 ; [OH
–
] = 10
–5
 = 2S
?
?
5
10
S
2
K
sp
 = [Ca
+2
] [OH
–
]
2
K
sp
 = S × (2S)
2
K
sp
 = 4S
3
K
sp
 = 
?
? ?
?
? ?
? ?
3
5
10
4
2
K
sp
= 0.5 × 10
–15
FINAL NEET(UG)–2019 EXAMINATION
(Held On Sunday 05
th
 MAY, 2019)
CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION
2
 
4. 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
Ans. (3)
Sol. N
2
(g) + 3H
2
(g) ??2NH
3
(g)
2 mole NH
3
(g) requires 3mole H
2
(g)
20 mole NH
3
(g) requires
= 
?
2
3
20 mole H (g)
2
= 30 mole
5. For an ideal solution, the correct option is :-
(1) ?
mix
 S = 0 at constant T and P (2) ?
mix
 V ? 0 at constant T and P
(3) ?
mix
 H = 0 at constant T and P (4) ?
mix
 G = 0 at constant T and P
Ans. (3)
Sol. For ideal solution ?H
mix
 = 0
6. For a cell involving one electron E
?
cell
 = 0.59V at 298 K, the equilibrium constant for the cell reaction is :-
? ?
? ?
? ?
? ?
2.303RT
Given that 0.059V at T 298K
F
(1) 1.0 × 10
2
(2) 1.0 × 10
5
(3) 1.0 × 10
10
(4) 1.0 × 10
30
Ans. (3)
Sol. E
cell
 = E
o
cell
 – 
10
2.303 RT
log Q
nF
at equlibrium E
cell
 = 0, Q = K
eq.
? ?
o
cell 10 eq.
0.0591
0 E log K
1
E
o
cell
 = +0.0591 log
10
 K
eq.
0.59 = + 0.0591 log
10
 K
eq.
+10 = log
10
 K
eq.
K
eq.
 = 10
+10
7. Among the following, the one that is not a green house gas is :-
(1) nitrous oxide (2) methane (3) ozone (4) sulphur dioxide
Ans. (4)
Sol. Besides carbon dioxide, other greenhouse gases are methane, water vapour, nitrous oxide, CFCs and ozone.
8. The number of sigma ( ?) and pi ( ?) bonds in pent-2-en-4-yne is :-
(1) 10 ? bonds and 3 ? bonds (2) 8 ? bonds and 5 ? bonds
(3) 11 ? bonds and 2 ? bonds (4) 13 ? bonds and no ? bond
Ans. (1)
Sol. H–C–C=C–C C–H ?
H
H H H
Number of sigma bonds = 10
Number of ?-bonds = 3
3
9. Which of the following diatomic molecular species has only ? bonds according to Molecular Orbital Theory ?
(1) O
2
(2) N
2
(3) C
2
(4) Be
2
Ans. (3)
Sol. According to M.O.T. electronic configuration of C
2
 molecule is -
?1s
2
 < ?*1s
2
 < ?2s
2
 < ?*2s
2
 < ?
2
p
x
2
= ?
2 
p
y
2
so, C
2
 molecule contain only ' ?' bond
10. Which of the following reactions are disproportionation reaction ?
(a) 2Cu
+
 ??Cu
2+
 + Cu
0
(b) 3MnO
4
2–
 + 4H
+
 ??2MnO
4
–
 + MnO
2
 + 2H
2
O
(c) 2KMnO
4
?
?? ?
 K
2
MnO
4
 + MnO
2
 + O
2
(d) 2MnO
4
–
 + 3Mn
2+
 + 2H
2
O ??5MnO
2
 + 4H
?
Select the correct option from the following :-
(1) (a) and (b) only (2) (a), (b) and (c) (3) (a), (c) and (d) (4) (a) and (d) only
Ans. (1)
Sol. (a) 2Cu
+
 ??Cu
+2
 + Cu
? ?
?
? ? ?
? ?
?
? ?
2
Cu Cu (oxidation)
Cu Cu (Reduction)
(b) MnO
4
2–
 ??MnO
4
–
 (oxidation)
+6          +7
MnO
4
2–
 ? ?MnO
2
 (Reduction)
+6            +4
The above two reaction are disproportionation.
11. Among the following, the narrow spectrum antibiotic is :-
(1) penicillin G (2) ampicillin (3) amoxycillin (4) chloramphenicol
Ans. (1)
Sol. The antibiotics which effective mainly against Gram-positive or Gram-negative bacteria are narrow spectrum
antibiotics. Penicillin G has a narrow spectrum.
ampicillin, amoxycillin, chloramphenicol are broad spectrum antibiotics.
12. The correct order of the basic strength of methyl substituted amines in aqueous solution is :-
(1) (CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N (2) (CH
3
)
3
N>CH
3
NH
2
 > (CH
3
)
2
NH
(3) (CH
3
)
3
N>(CH
3
)
2
NH>CH
3
NH
2
(4) CH
3
NH
2
>(CH
3
)
2
NH > (CH
3
)
3
N
Ans. (1)
Sol. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution
is as follows :
(C
2
H
5
)
2
 NH > (C
2
H
5
)
3
N > C
2
H
5
NH
2
 >NH
3
(CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N > NH
3
13. Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI] I
–
 sol. ?
(1) 50 mL of 1M AgNO
3
 + 50 mL of 1.5 M KI
(2) 50 mL of 1M 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
Ans. (1,2)
Sol. In negatively charged colloid [AgI] I
–
, I
–
 is preferentially adsorbed.
AgNO
3
 + KI ??AgI + KNO
3
When KI is in excess, I
–
 will be adsorbed on the surface of AgI and [AgI] I
–
 is formed
4
 
14. Conjugate base for Bronsted acids H
2
O and HF are:-
(1) OH
–
 and H
2
F
+
 respectively (2) H
3
O
+
 and F
–
, respectively
(3) OH
–
 and F
–
, respectively (4) H
3
O
+
 and H
2
F
+
, respectively
Ans. (3)
Sol. Conjugate base of H
2
O is OH
–
Conjugate base of HF is F
–
15. Which will make basic buffer ?
(1) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH
3
COOH
(2) 100 mL of 0.1 M CH
3
COOH + 100 mL of 0.1M NaOH
(3) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH
4
OH
(4) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Ans. (3)
Sol. Basic buffer is mixture of weak base and salt of weak base with strong acid
milli mole of HCl = 100 × 0.1 = 10 milli mole
milli mole of NH
4
OH = 200 × 0.1 = 20 milli mole
HCl + NH
4
OH ??NH
4
Cl + H
2
O
 10      20          –         –
 –       10           10
16. The compound that is most difficult to protonate is:-
(1) 
H
O
H
(2) 
H C
3
O
H
(3) 
H C
3
O
CH
3
(4) 
Ph
O
H
Ans. (4)
Sol. In case of phenol  lone pair of oxygen is delocalized in ring.
O–H
:
:
17. The most suitable reagent for the following conversion is :-
H C–C C–CH
3 3
?
H C
3
H
CH
3
H
cis-2-butene
(1) Na/liquid NH
3
(2) H
2
, Pd/C, quinoline
(3) Zn/HCl (4) Hg
2+
/H
+
, H
2
O
Ans. (2)
Sol.
H C–C C–CH
3 3
?
H C
3
H
CH
3
H
H ,Pd/C
2
quinoline
C=C
(syn-addition)
18. Which of the following species is not stable ?
(1) [SiF
6
]
2–
(2) [GeCl
6
]
2–
(3) [Sn(OH)
6
]
2–
(4) [SiCl
6
]
2–
Ans. (4)
Sol.
? 2
6
SiCl does not exist since
(i) size of Cl
–
 is large so it cannot accommodate around Si
+4
 due to limitation of size
(ii) Interaction between lone pair of chloride ion and Si
+4
 is not very strong
Page 5


1
1. Under isothermal condition, a gas at 300 K expands from 0.1L to 0.25L against a constant external pressure
of 2 bar. The work done by the gas is :-
[Given that 1L bar = 100 J]
(1) –30 J (2) 5kJ (3) 25 J (4) 30 J
Ans. (1)
Sol. W = –P
ext
 (V
2
–V
1
)
P
ext 
 = 2 bar
V
1
 = 0.1 L
V
2
 = 0.25 L
W = –2 bar[0.25 – 0.1] L
W = –2 × 0.15 bar L
W = –0.30 bar L
W = (–0.30) × 100 = –30 J
2. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and
the cations occupy 75% of octahedral voids. The formula of the compound is :-
(1) C
2
A
3
(2) C
3
A
2
(3) C
3
A
4
(4) C
4
A
3
Ans. (3)
Sol. Anion A in HCP
No of ions of A in Unit cell = 6
No of Octahedral voids = 6
75% is occupied by cations C
No of cations C = ?
75
6
100
= ?
3
6
4
= 
9
2
C
9/2
A
6
C
9
A
12
Simple ratio C
3
A
4
3. 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
Ans. (1)
Sol. Ca(OH)
2
(s) ??Ca
+2
(aq) + 2OH
–
(aq)
   S            2S
pH = 9 ;  pOH = 5 ; [OH
–
] = 10
–5
 = 2S
?
?
5
10
S
2
K
sp
 = [Ca
+2
] [OH
–
]
2
K
sp
 = S × (2S)
2
K
sp
 = 4S
3
K
sp
 = 
?
? ?
?
? ?
? ?
3
5
10
4
2
K
sp
= 0.5 × 10
–15
FINAL NEET(UG)–2019 EXAMINATION
(Held On Sunday 05
th
 MAY, 2019)
CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION
2
 
4. 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
Ans. (3)
Sol. N
2
(g) + 3H
2
(g) ??2NH
3
(g)
2 mole NH
3
(g) requires 3mole H
2
(g)
20 mole NH
3
(g) requires
= 
?
2
3
20 mole H (g)
2
= 30 mole
5. For an ideal solution, the correct option is :-
(1) ?
mix
 S = 0 at constant T and P (2) ?
mix
 V ? 0 at constant T and P
(3) ?
mix
 H = 0 at constant T and P (4) ?
mix
 G = 0 at constant T and P
Ans. (3)
Sol. For ideal solution ?H
mix
 = 0
6. For a cell involving one electron E
?
cell
 = 0.59V at 298 K, the equilibrium constant for the cell reaction is :-
? ?
? ?
? ?
? ?
2.303RT
Given that 0.059V at T 298K
F
(1) 1.0 × 10
2
(2) 1.0 × 10
5
(3) 1.0 × 10
10
(4) 1.0 × 10
30
Ans. (3)
Sol. E
cell
 = E
o
cell
 – 
10
2.303 RT
log Q
nF
at equlibrium E
cell
 = 0, Q = K
eq.
? ?
o
cell 10 eq.
0.0591
0 E log K
1
E
o
cell
 = +0.0591 log
10
 K
eq.
0.59 = + 0.0591 log
10
 K
eq.
+10 = log
10
 K
eq.
K
eq.
 = 10
+10
7. Among the following, the one that is not a green house gas is :-
(1) nitrous oxide (2) methane (3) ozone (4) sulphur dioxide
Ans. (4)
Sol. Besides carbon dioxide, other greenhouse gases are methane, water vapour, nitrous oxide, CFCs and ozone.
8. The number of sigma ( ?) and pi ( ?) bonds in pent-2-en-4-yne is :-
(1) 10 ? bonds and 3 ? bonds (2) 8 ? bonds and 5 ? bonds
(3) 11 ? bonds and 2 ? bonds (4) 13 ? bonds and no ? bond
Ans. (1)
Sol. H–C–C=C–C C–H ?
H
H H H
Number of sigma bonds = 10
Number of ?-bonds = 3
3
9. Which of the following diatomic molecular species has only ? bonds according to Molecular Orbital Theory ?
(1) O
2
(2) N
2
(3) C
2
(4) Be
2
Ans. (3)
Sol. According to M.O.T. electronic configuration of C
2
 molecule is -
?1s
2
 < ?*1s
2
 < ?2s
2
 < ?*2s
2
 < ?
2
p
x
2
= ?
2 
p
y
2
so, C
2
 molecule contain only ' ?' bond
10. Which of the following reactions are disproportionation reaction ?
(a) 2Cu
+
 ??Cu
2+
 + Cu
0
(b) 3MnO
4
2–
 + 4H
+
 ??2MnO
4
–
 + MnO
2
 + 2H
2
O
(c) 2KMnO
4
?
?? ?
 K
2
MnO
4
 + MnO
2
 + O
2
(d) 2MnO
4
–
 + 3Mn
2+
 + 2H
2
O ??5MnO
2
 + 4H
?
Select the correct option from the following :-
(1) (a) and (b) only (2) (a), (b) and (c) (3) (a), (c) and (d) (4) (a) and (d) only
Ans. (1)
Sol. (a) 2Cu
+
 ??Cu
+2
 + Cu
? ?
?
? ? ?
? ?
?
? ?
2
Cu Cu (oxidation)
Cu Cu (Reduction)
(b) MnO
4
2–
 ??MnO
4
–
 (oxidation)
+6          +7
MnO
4
2–
 ? ?MnO
2
 (Reduction)
+6            +4
The above two reaction are disproportionation.
11. Among the following, the narrow spectrum antibiotic is :-
(1) penicillin G (2) ampicillin (3) amoxycillin (4) chloramphenicol
Ans. (1)
Sol. The antibiotics which effective mainly against Gram-positive or Gram-negative bacteria are narrow spectrum
antibiotics. Penicillin G has a narrow spectrum.
ampicillin, amoxycillin, chloramphenicol are broad spectrum antibiotics.
12. The correct order of the basic strength of methyl substituted amines in aqueous solution is :-
(1) (CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N (2) (CH
3
)
3
N>CH
3
NH
2
 > (CH
3
)
2
NH
(3) (CH
3
)
3
N>(CH
3
)
2
NH>CH
3
NH
2
(4) CH
3
NH
2
>(CH
3
)
2
NH > (CH
3
)
3
N
Ans. (1)
Sol. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution
is as follows :
(C
2
H
5
)
2
 NH > (C
2
H
5
)
3
N > C
2
H
5
NH
2
 >NH
3
(CH
3
)
2
NH > CH
3
NH
2
 > (CH
3
)
3
N > NH
3
13. Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI] I
–
 sol. ?
(1) 50 mL of 1M AgNO
3
 + 50 mL of 1.5 M KI
(2) 50 mL of 1M 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
Ans. (1,2)
Sol. In negatively charged colloid [AgI] I
–
, I
–
 is preferentially adsorbed.
AgNO
3
 + KI ??AgI + KNO
3
When KI is in excess, I
–
 will be adsorbed on the surface of AgI and [AgI] I
–
 is formed
4
 
14. Conjugate base for Bronsted acids H
2
O and HF are:-
(1) OH
–
 and H
2
F
+
 respectively (2) H
3
O
+
 and F
–
, respectively
(3) OH
–
 and F
–
, respectively (4) H
3
O
+
 and H
2
F
+
, respectively
Ans. (3)
Sol. Conjugate base of H
2
O is OH
–
Conjugate base of HF is F
–
15. Which will make basic buffer ?
(1) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH
3
COOH
(2) 100 mL of 0.1 M CH
3
COOH + 100 mL of 0.1M NaOH
(3) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH
4
OH
(4) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Ans. (3)
Sol. Basic buffer is mixture of weak base and salt of weak base with strong acid
milli mole of HCl = 100 × 0.1 = 10 milli mole
milli mole of NH
4
OH = 200 × 0.1 = 20 milli mole
HCl + NH
4
OH ??NH
4
Cl + H
2
O
 10      20          –         –
 –       10           10
16. The compound that is most difficult to protonate is:-
(1) 
H
O
H
(2) 
H C
3
O
H
(3) 
H C
3
O
CH
3
(4) 
Ph
O
H
Ans. (4)
Sol. In case of phenol  lone pair of oxygen is delocalized in ring.
O–H
:
:
17. The most suitable reagent for the following conversion is :-
H C–C C–CH
3 3
?
H C
3
H
CH
3
H
cis-2-butene
(1) Na/liquid NH
3
(2) H
2
, Pd/C, quinoline
(3) Zn/HCl (4) Hg
2+
/H
+
, H
2
O
Ans. (2)
Sol.
H C–C C–CH
3 3
?
H C
3
H
CH
3
H
H ,Pd/C
2
quinoline
C=C
(syn-addition)
18. Which of the following species is not stable ?
(1) [SiF
6
]
2–
(2) [GeCl
6
]
2–
(3) [Sn(OH)
6
]
2–
(4) [SiCl
6
]
2–
Ans. (4)
Sol.
? 2
6
SiCl does not exist since
(i) size of Cl
–
 is large so it cannot accommodate around Si
+4
 due to limitation of size
(ii) Interaction between lone pair of chloride ion and Si
+4
 is not very strong
5
19. Which of the following is an amphoteric hydroxide?
(1) Sr(OH)
2
(2) Ca(OH)
2
(3) Mg(OH)
2
(4) Be(OH)
2
Ans. (4)
Sol. Be(OH)
2
 is an amphoteric hydroxide rest all are basic hydroxides
20. The structure of intermediate A in the following reaction is :-
O
2
CH
CH
3
CH
3
A
H
+
H O
2
OH
+
O
H C
3
CH
3
(1)
O
CH
CH
3
CH
3
(2) 
H C–C–O–O–H
3
CH
3
(3)  
O–O–CH
CH
3
CH
3
(4)  
HC
CH –O–O–H
2
CH
3
Ans. (2)
Sol. Phenol is manufactured from the hydrocarbon, cumene. Cumene (isopropylbenzene) is oxidised in the presence
of air to cumene hydroperoxide. it is converted to phenol and acetone by treating it with dilute acid. Acetone,
a by-product of this reaction, is also obtained in large quantities by this method.
  
CH –CH
3
CH
3
Cumene
O
2
CH –C–O–O–H
3
CH
3
Cumene
hydroperoxide
H
+
H O
2
OH
  +  CH COCH
3 3
21. The manganate and permanganate ions are tetrahedral, due to
(1) The ?–bonding involves overlap of p–orbitals of oxygen with d–orbitals of manganese
(2) There is no ?–bonding
(3) The ?–bonding involves overlap of p–orbitals of oxygen with p–orbitals of managanese
(4) The ?–bonding involves overlap of d–orbitals of oxygen with d–orbitals of manganese
Ans. (1)
Sol.
?2
4
MnO (Mangnate ion) and 
?
4
MnO (Permangnate ion)
both are tetrahedral
Mn
O
O
O
O
Mn
O
O
O
O
Since ' ?' bond is formed between p-orbital of oxygen and d–orbital of Managnese
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