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Byju's Answer
Standard XII
Chemistry
Equivalent Mass
50 ml of 0....
Question
50
ml of
0.2
N
K
2
C
r
2
O
7
is required to oxidize
40
ml of
F
e
S
O
4
solution in an acidic medium.
The weight of iron present in
1
L
of
F
e
S
O
4
solution is (atomic weight of iron is
56
):
A
14
g
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B
1.4
g
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C
140
g
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D
28
g
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Solution
The correct option is
B
14
g
The valency factor for
K
2
C
r
2
O
7
is 6.
N
1
V
1
=
N
2
V
2
⟹
0.2
×
50
=
N
2
×
40
⟹
N
2
=
10
40
=
1
4
⟹
N
2
=
w
(
M
.
w
t
)
×
e
q
.
f
a
c
t
o
r
×
1
V
(
l
i
t
)
⟹
1
4
=
w
56
×
1
×
1
⟹
w
=
56
4
⟹
w
=
14
g
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1
Similar questions
Q.
50 mL aqueous solution of
F
e
S
O
4
required 12 mL of 0.2 M
K
M
n
O
4
in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:
M
n
O
−
4
+
5
F
e
2
+
+
8
H
+
→
M
n
2
+
+
5
F
e
3
+
+
4
H
2
O
Q.
A
8
g
m
sample of iron are containing
V
%
(by wt) of iron
F
e
2
+
was present in
1
L
solution. This solution required
V
m
L
of a
K
M
n
O
4
solution for oxidation of
F
e
2
+
in acidic medium. If in another titration
20
m
L
of same
K
M
n
O
4
is required, for oxidation of
20
m
L
H
2
O
2
in acidic medium. Find volume strength of
H
2
O
2
Q.
3
moles of a mixture of
F
e
S
O
4
and
F
e
2
(
S
O
4
)
3
required
100
mL of
2
M
K
M
n
O
4
solution in acidic medium. Hence, mole fraction of
F
e
S
O
4
in the mixture is
:
Q.
How many grams of
K
2
C
r
2
O
7
is required to oxidise
12.41
g of
F
e
S
O
4
to
F
e
2
(
S
O
4
)
3
in an acidic medium?
(
K
2
C
r
2
O
7
=
294
;
F
e
S
O
4
=
152
)
Q.
The weight of
K
M
n
O
4
required to completely oxidise
0.25
moles of
F
e
S
O
4
in acidic medium is:
[Molecular weight of
K
M
n
O
4
=
158
]
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