Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
3.92 g of ferrous ammonium sulphate are dissolved in 100 mL water. 20 mL of this solution requires 18 mL of potassium permanganate during titration for complete oxidation. The weight of KMnO 4 present in one litre of the solution is
Q.
3.92
g
of ferrous ammonium sulphate are dissolved in
100
m
L
water.
20
m
L
of this solution requires
18
m
L
of potassium permanganate during titration for complete oxidation. The weight of
K
M
n
O
4
present in one litre of the solution is
2850
253
Redox Reactions
Report Error
A
34.76
g
0%
B
12.38
g
0%
C
1.238
g
0%
D
3.51
g
100%
Solution:
The redox reaction involving the oxidation of
F
e
2
+
(from ferrous ammonium sulphate) is
M
n
O
4
−
+
5
F
e
2
+
+
8
H
+
⟶
5
F
e
3
+
+
M
n
2
+
+
4
H
2
O
Mol. wt. of ferrous ammonium sulphate,
(
N
H
4
)
2
S
O
4
⋅
F
e
S
O
4
⋅
6
H
2
O
=
392
∴
Molarity of
(
N
H
4
)
2
S
O
4
⋅
F
e
S
O
4
⋅
6
H
2
O
=
Mol.
wt.
W
t
×
Volume
1000
=
392
3.92
×
100
1000
=
0.1
Applying molarity equation,
n
1
M
1
V
1
(Ferrous amm. sulphate)
=
n
2
M
2
V
2
(
K
M
n
O
4
)
or
5
0
⋅
1
×
20
=
1
M
2
×
18
or
M
2
=
5
×
18
0
⋅
1
×
20
=
45
1
Amount of
K
M
n
O
4
present in one litre
=
Molarity
×
Mol. wt.
=
45
1
×
158
=
3
⋅
51
g