Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
A 100 % pure sample of a divalent metal carbonate weighing 2 g on complete thermal decomposition releases 448 cc of carbon dioxide at STP. The equivalent mass of the metal is
Q. A 100 % pure sample of a divalent metal carbonate weighing
2
g
on complete thermal decomposition releases
448
cc
of carbon dioxide at
STP
. The equivalent mass of the metal is
2841
199
KEAM
KEAM 2012
Some Basic Concepts of Chemistry
Report Error
A
40
0%
B
20
71%
C
28
0%
D
12
0%
E
56
0%
Solution:
Let the metal is
M
, so the formula of its carbonate is
MC
O
3
.
Molar mass of
MC
O
3
=
x
+
12
+
3
×
16
=
(
x
+
60
)
g
/
m
o
l
(Let atomic mass of
M
is
x
.)
1
m
o
l
MC
O
3
→
Δ
MO
+
1
m
o
l
C
O
2
=
22.4
L
=
22400
M
L
∵
448
cc
(
448
m
L
)
C
O
2
is produced from carbonate
=
2
g
∴
22400
cc
C
O
2
will be obtained from carbonate
=
448
2
×
22400
=
100
g
∴
100
=
x
+
60
x
=
100
−
60
=
40
g
/
m
o
l
Eq. wt. of metal
=
2
40
=
20
g
equiv
−
1