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Byju's Answer
Standard XII
Chemistry
Common Ion Effect
Weight of N...
Question
Weight of
N
a
2
C
O
3
which can be neutralised by 100 ml of 0.05M
H
2
S
O
4
is:
A
0.53
g
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B
1.06
g
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C
5.3
g
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D
10.6
g
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Solution
The correct option is
A
0.53
g
Using stoichiometry,
E
q
H
2
S
O
4
=
E
q
N
a
2
C
O
3
2
×
0.05
×
100
1000
=
2
×
m
o
l
e
s
(
N
a
2
C
O
3
)
M
o
l
e
s
o
f
N
a
2
C
O
3
=
5
1000
Therefore, weight of
N
a
2
C
O
3
=
106
×
5
1000
g
m
=
0.53
g
m
Suggest Corrections
0
Similar questions
Q.
100
m
l
of
N
a
2
C
O
3
solution containing
5.3
g
of
N
a
2
C
O
3
was exactly neutralised by
200
m
l
of
H
2
S
O
4
solution. What is the pH of
H
2
S
O
4
?
(
l
o
g
5
=
0.6990
)
Q.
1.25
g
of a sample of
N
a
2
C
O
3
and
N
a
2
S
O
4
is dissolved in
250
m
l
solution.
25
m
l
of this solution neutralises
20
m
l
of
0.1
N
H
2
S
O
4
. The % of
N
a
2
C
O
3
in this sample is:
Q.
A solution contain
N
a
2
C
O
3
and
N
a
H
C
O
3
. 10 ml of this solution required 2.5 ml of 0.1 M
H
2
S
O
4
for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M
H
2
S
O
4
was required. The amount of
N
a
2
C
O
3
and
N
a
H
C
O
3
in 1 litre of the solution is ?
Q.
25
g
m
mixture of
N
a
2
C
O
3
and
N
a
2
S
O
4
are dissolved in
250
m
l
of solution.
25
m
l
of above solution required
20
m
l
of
0.1
N
H
2
S
O
4
using phenolphthalein indicator. The mass of
N
a
2
C
O
3
in the mixture is:
Q.
250
m
L
of a
N
a
2
C
O
3
solution contains
2.65
g
of
N
a
2
C
O
3
⋅
10
m
L
of this solution is added to
x
m
L
of water to obtain
0.001
M
N
a
2
C
O
3
solution. The value of
x
is:
[Molecular weight of
N
a
2
C
O
3
=
106
]
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