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JEE Main Chemistry Class 12 Analytical Chemistry Questions
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© examsnet.com
Question : 66
Total: 106
An organic compound '
A
' is oxidized with
N
a
2
O
2
followed by boiling with
H
N
O
3
. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is:
[April 12, 2019 (I)]
Nitrogen
Phosphorus
Fluorine
Sulphur
Validate
Solution:
Phosphorus is detected in the form of yellow ppt of ammonium phosphate molybdate on reaction with ammonium molybdate.
2
P
(
A
)
+
3
N
a
2
O
2
+
O
2
fusion
─
─
─
─
─
─
▶
2
N
a
3
P
O
4
N
a
3
P
O
4
+
3
H
N
O
3
─
─
─
─
─
─
▶
H
3
P
O
4
+
3
N
a
N
O
3
H
3
P
O
4
+
12
(
N
H
4
)
2
M
0
O
4
+
21
H
N
O
3
(
Ammoniun
molybdate
)
─
─
─
─
─
─
▶
(
N
H
4
)
3
P
O
4
⋅
12
M
0
O
3
↓
+
21
N
H
4
N
O
3
+
12
H
2
O
(
Ammonium
phosphomolybdate
)
(
canary
yellow
ppt
.
)
© examsnet.com
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