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Byju's Answer
Standard XII
Chemistry
Molality
Presence of ...
Question
Presence of
C
a
S
O
4
causes permanent hardness in water. It can be removed by precipitating it as
C
a
C
O
3
by adding
N
a
2
C
O
3
. If the amount of
N
a
2
C
O
3
required to soften 100 L of hard water, which is 200 ppm
C
a
C
O
3
, is
x
g, then value of
x
is:
Open in App
Solution
N
a
2
C
O
3
106
g
+
C
a
S
O
4
→
C
a
C
O
3
100
g
↓
+
N
a
2
S
O
4
10
6
m
L
o
f
h
a
r
d
w
a
t
e
r
h
a
s
=
200
g
C
a
C
O
3
100
L
=
10
5
m
L
o
f
h
a
r
d
w
a
t
e
r
h
a
s
=
20
g
C
a
C
O
3
I
f
h
a
r
d
n
e
s
s
i
s
100
g
,
N
a
2
C
O
3
r
e
q
u
i
r
e
d
=
106
g
I
f
h
a
r
d
n
e
s
s
i
s
20
g
,
N
a
2
C
O
3
r
e
q
u
i
r
e
d
=
21.2
g
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0
Similar questions
Q.
Permanent hardness is due to
S
O
2
−
4
and
C
l
−
of
C
a
2
+
and
M
g
2
+
and is removed by the addition of
N
a
2
C
O
3
.
C
a
S
O
4
+
N
a
2
C
O
3
⟶
C
a
C
O
3
+
N
a
2
S
O
4
C
a
C
l
2
+
N
a
2
C
O
3
⟶
C
a
C
O
3
+
2
N
a
C
l
If hardness is
100
ppm
C
a
C
O
3
, amount of
N
a
2
C
O
3
required to soften
10
L of hard water is
:
Q.
Permanent hardness is due to
C
l
⊝
and
S
O
2
−
4
of
M
g
2
+
and
C
a
2
+
and is removed by adding
N
a
2
C
O
3
.
C
a
S
O
4
+
N
a
2
C
O
3
→
C
a
C
O
3
+
N
a
2
S
O
4
C
a
C
l
2
+
N
a
2
C
O
3
→
C
a
C
O
3
+
2
N
a
C
l
Which of the following statements is/are correct?
Q.
N
a
2
C
O
3
is widely used in softening of hard water. If 1 L of hard water required
0.0106
g
of
N
a
2
C
O
3
, The hardness in ppm (parts per million i.e.,
10
6
ml) of
C
a
C
O
3
is:
Q.
Following reaction is used to remove temporary hardness which is due to dissolved bicarbonates of
C
a
2
+
and
M
g
2
+
.
C
a
(
H
C
O
3
)
2
+
C
a
(
O
H
)
2
→
2
C
a
C
O
3
↓
+
2
H
2
O
Actually lime (
C
a
O
) is added which in
H
2
O
forms
C
a
(
O
H
)
2
and precipitates
C
a
C
O
3
C
a
O
+
H
2
O
→
C
a
(
O
H
)
2
If
10
L of hard water requires
0.56
g of
C
a
O
, temporary hardness of
C
a
C
O
3
is
x
×
10
2
ppm. Find
x
.
Q.
Hardness of water is measured in terms of ppm (parts per million) of
C
a
C
O
3
. It is the amount (in g) of
C
a
C
O
3
present in
10
6
g
H
2
O
. In a sample of water,
10
L required
0.56
g of
C
a
O
to the remove temporary hardness of
H
C
O
−
3
.
C
a
(
H
C
O
3
)
2
+
C
a
O
⟶
2
C
a
C
O
3
+
H
2
O
Temporary hardness is
:
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