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Class 11 NEET Chemistry Some Basic Concepts Of Chemistry
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© examsnet.com
Question : 12
Total: 63
What mass of
95
%
pure
C
a
C
O
3
will be required to neutralise
50
m
L
of
0.5
M
H
C
l
solution according to the following reaction?
C
a
C
O
3
(
s
)
+
2
H
C
l
‌
(aq)
‌
→
C
a
C
l
2
(
‌
aq
‌
)
+
C
O
2
(
‌
g
‌
)
+
2
H
2
O
‌
(l)
‌
[Calculate upto second place of decimal point]
[NEET-2022]
1.25
g
1.32
g
3.65
g
9.50
g
Validate
Solution:
👈: Video Solution
Let
m
gram mass of
C
a
C
O
3
is required
Pure
C
a
C
O
3
in
m
gram
=
‌
95
100
×
m
Moles of
C
a
C
O
3
=
‌
95
100
×
‌
m
100
Moles of
H
C
l
required
=
2
×
moles of
C
a
C
O
3
‌
‌
=
2
×
‌
95
100
×
‌
m
100
2
×
‌
95
100
×
‌
m
100
=
‌
50
1000
×
0.5
m
=
1.315
g
≈
1.32
g
© examsnet.com
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