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JEE Main 17 March 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 54
Total: 90
In the above reaction
3.9
g
of benzene on nitration gives
4.92
g
of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ......... %
(Round off to the nearest integer). (Given, atomic mass
C
:
12.0
u
,
H
:
1.0
u
,
O
:
16.0
u
,
N
:
14.0
u
)
[17 Mar 2021 Shift 1]
Your Answer:
Validate
Solution:
C
6
H
6
H
N
O
3
─
─
─
─
▶
H
2
S
O
4
C
6
H
5
N
O
2
( Molar mass
=
123
)
Moles of
C
6
H
6
=
(
3.9
78
)
=
0.05
So, moles of
C
6
H
5
N
O
2
formed should be
3.9
78
.
Mass of
C
6
H
5
N
O
2
=
Moles
×
Molar mass
=
3.9
78
×
123
=
6.15
g
By conserving moles of carbon, mole or
C
6
H
5
N
O
2
.
%
yield
=
Weight actually formed
Weight theoretical
×
100
=
4.92
6.15
×
100
=
80
%
© examsnet.com
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