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Byju's Answer
Standard XII
Chemistry
Concentration Terms - An Introduction (w/w etc)
Cr2O-27 + I- ...
Question
C
r
2
O
−
2
7
+
I
−
+
H
+
→
C
r
+
3
+
I
2
+
H
2
O
The equivalent weight of the reductant in the above equation is: (At . wt. of Cr = 52, I = 127)
A
26
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B
127
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C
63.5
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D
10.4
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Solution
The correct option is
A
127
This reaction is disproportionate reaction
n
f
a
c
t
o
r
=
n
1
×
n
2
n
1
+
n
2
=
6
×
1
6
+
1
−
6
7
Equilateral weight
=
m
o
l
a
r
m
a
s
s
n
f
Reductase here is
I
−
Equilateral weight
=
127
1
=
127
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Similar questions
Q.
C
2
O
−
2
7
+
I
−
+
H
+
→
C
r
+
3
+
l
2
+
H
2
O
The equivalent weight of the reductant in the above equation is :-
(
A
t
.
w
t
.
of
C
r
=
52
,
I
=
127
)
Q.
What is the increasing order of equivalent weights of the following oxidising agents?
(i)
K
M
n
O
4
⟶
M
n
2
+
(ii)
K
2
C
r
2
O
7
⟶
C
r
3
+
(iii)
K
C
l
O
3
⟶
C
l
−
[Atomic wt.:
K
=
39
,
C
r
=
52
,
M
n
=
55
,
C
l
=
35.5
,
O
=
16
]
Q.
Balance by Ion electon method,
C
r
2
O
−
2
7
+
F
e
+
2
→
C
r
+
3
+
H
2
O
(in acidic medium).
Q.
C
r
2
O
2
−
7
H
+
⟶
C
r
3
+
Find the equivalent weight of
C
r
2
O
2
−
7
in the above redox half reaction.
(Given the molar mass of
C
r
2
O
2
−
7
is
M
g
/
m
o
l
)
Q.
In an oxidation-reduction reaction, dichromate
(
C
r
2
O
−
2
7
)
ion is reduced to
C
r
+
3
ion. The equivalent weight of
K
2
C
r
2
O
7
in this reaction is:
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