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AP EAMCET Engineering 2011 Solved Paper
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© examsnet.com
Question : 78
Total: 160
19
g
of a mixture containing
N
a
H
C
O
3
and
N
a
2
C
O
3
on complete heating liberated
1.12
L
of
C
O
2
at STP. The weight of the remaining solid was
15.9
g
. What is the weight (in
g
) of
N
a
2
C
O
3
in the mixture before heating?
8.4
15.9
4.0
10.6
Validate
Solution:
Molecular weight of
N
a
H
C
O
3
=
23
+
1
+
12
+
48
=
84
Molecular weight of
N
a
2
C
O
3
=
46
+
12
+
48
=
106
Hence, total weight
=
84
+
106
=
190
∵
In
190
g
of a mixture, weight of
N
a
2
C
O
3
is
=
106
∴
In
19
g
of a mixture weight of
N
a
2
C
O
3
=
106
×
19
190
=
10.6
g
© examsnet.com
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