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Standard XII
Chemistry
Introduction to Oxidation and Reduction
Equivalent ma...
Question
Equivalent mass of
K
M
n
O
4
in acidic, basic and neutral are in the ratio of:
A
3
:
5
:
15
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B
5
:
3
:
1
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C
5
:
1
:
3
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D
3
:
15
:
5
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Solution
The correct option is
D
3
:
15
:
5
In acidic medium:
M
n
O
−
4
+
8
H
+
+
5
e
−
→
M
n
+
2
+
4
H
2
O
In basic medium:
M
n
O
−
4
+
e
−
→
M
n
O
2
−
4
In neutral medium:
M
n
O
−
4
+
4
H
+
+
3
e
−
→
M
n
O
2
+
2
H
2
O
Equivalent mass is molecular mass divided by number of electrons gained.
In acidic medium, equivalent weight
=
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
5
In basic medium, equivalent weight
=
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
1
In neutal medium, equivalent weight
=
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
3
Ration of equivalent mass
=
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
5
:
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
1
:
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
3
Ratio of equivalent mass
=
3
:
15
:
5
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