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Byju's Answer
Standard XII
Chemistry
Percentage Composition
27.8 g of F...
Question
27.8
g of
F
e
S
O
4
.
x
H
2
O
is treated with excess of
B
a
C
l
2
to give
2.33
g of white precipitate. Calculate the value of
x
.
(
B
a
=
137
,
S
=
32
,
F
e
=
56
,
H
=
1
,
O
=
16
)
A
7
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B
6
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C
5
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D
8
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Solution
The correct option is
B
7
F
e
S
O
4
.
x
H
2
O
+
B
a
C
l
2
→
B
a
S
O
4
1
m
o
l
e
1
m
o
l
e
2.78
g
m
2.33
g
m
96
+
56
+
18
x
=
137
+
96
So,
x
=
7
.
Suggest Corrections
0
Similar questions
Q.
10
mL of a solution of
K
2
S
O
4
, on reaction with
B
a
C
l
2
, gave
0.233
g of white ppt. The concentration of
K
2
S
O
4
solution is
:
(Given that the molecular weight of
B
a
=
137
,
S
=
32
,
O
=
16
,
K
=
39
)
Q.
500
mL of
0.25
M
N
a
2
S
O
4
solution is added to an aqueous solution of
15.0
g of
B
a
C
l
2
solution resulting in the formation of white precipitate of
B
a
S
O
4
. The moles of
B
a
S
O
4
formed is
y
×
10
−
3
moles. Then,
y
is (in nearest integer) : [Given that molecular weight of
B
a
=
137
,
S
=
32
,
O
=
16
,
N
a
=
23
g/mol]
Q.
The equivalent weight of
F
e
S
O
4
.
(
N
H
4
)
2
S
O
4
.24
H
2
O
is:
(At. wt of
F
e
=
56
,
S
=
32
,
O
=
16
,
N
=
14
,
H
=
1
)
Q.
The amount of
B
a
S
O
4
formed upon mixing
100
ml of
20.8
%
B
a
C
l
2
solution with
50
ml of
9.8
%
H
2
S
O
4
solution will be:
(Given that molecular weight of
B
a
=
137
,
C
l
=
35.5
,
S
=
32
,
H
=
1
and
O
=
16
g/mol)
Q.
100 mL of 20.8%
B
a
C
l
2
solution and 50 mL of 9.8%
H
2
S
O
4
solution will form
B
a
S
O
4
(
B
a
=
137
,
C
l
=
35.5
,
S
=
32
,
H
=
1
,
O
=
16
)
B
a
C
l
2
+
H
2
S
O
4
→
B
a
S
O
4
+
2
H
C
l
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