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Question
Chemistry
10 L of hard water required 0.56 g of lime ( CaO ) for removing hardness Ca ( HCO 3)2+ CaO -2 CaCO 3+ H 2 O Thus, temporary hardness in terms of ppm of CaCO 3 is
Q.
10
L
of hard water required
0.56
g
of lime
(
C
a
O
)
for removing hardness
C
a
(
H
C
O
3
)
2
+
C
a
O
−
2
C
a
C
O
3
+
H
2
O
Thus, temporary hardness in terms of
pp
m
of
C
a
C
O
3
is
1816
244
Some Basic Concepts of Chemistry
Report Error
A
10
B
20
C
100
D
200
Solution:
C
a
(
H
C
O
3
)
2
+
C
a
O
1
m
o
l
56
g
0.56
g
→
2
C
a
C
O
3
+
H
2
O
2
m
o
l
200
g
2.00
g
Thus,
10
L
water
(
=
1
0
4
g
)
water has hardness
=
2
g
∴
1
0
6
g
(
pp
m
)
water has hardness
=
1
0
4
2
×
1
0
6
=
200