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Byju's Answer
Standard XII
Chemistry
Sucrose
Excess of aci...
Question
Excess of acidified solution of KI was added to 30mL H2O2 solution. The iodine liberated required 25mL of 0.4 N Na2S2O3 solution. The volume strength of H2O2 solution is?
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Solution
T
h
e
t
i
t
r
a
t
i
o
n
w
a
s
d
o
n
e
i
n
d
i
r
e
c
t
l
y
w
i
t
h
H
y
d
r
o
g
e
n
p
e
r
o
x
i
d
e
a
n
d
t
h
i
o
s
o
l
u
t
i
o
n
u
sin
g
K
I
.
V
1
N
1
=
V
2
N
2
30
×
N
1
=
25
×
0
.
4
N
1
=
25
×
0
.
4
30
=
0
.
333
N
V
o
l
u
m
e
s
t
r
e
n
g
t
h
=
N
o
r
m
a
l
i
t
y
×
5
.
6
=
0
.
333
×
5
.
6
=
1
.
867
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0
Similar questions
Q.
To a 25 ml
H
2
O
2
solution, excess of acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N
N
a
2
S
2
O
3
solution. The volume strength of
H
2
O
2
solution is:
Q.
To a
25
m
l
H
2
O
2
solution, excess of acidified solution of
K
I
was added. The iodine liberated required
20
m
l
of
0.3
N
N
a
2
S
2
O
3
solution. The volume strength of
H
2
O
2
solution is:
Q.
An excess of acidified solution of
K
I
is added to a
25
mL
H
2
O
2
solution. The iodine liberated requires
20
mL of
0.3
N
N
a
2
S
2
O
3
solution. Calculate the volume strength of
H
2
O
2
:
Q.
25
ml of
H
2
O
2
solution were added to excess of acidified
K
I
solution. The iodine so liberated required
20
ml of
0.1
N
N
a
2
S
2
O
3
for titration. Calculate strength of
H
2
O
2
in terms of normality and percentage?
Q.
To a
25
mL
H
2
O
2
solution, excess of an acidified solution of potassium iodide was added. The iodine liberated required
20
mL of
0.3
N sodium thiosulphate solution.
Calculate the volume strength of
H
2
O
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solution (to the nearest integer):
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