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Byju's Answer
Standard XII
Chemistry
Laws of Mass Conservation
4 g equimolar...
Question
4 g equimolar mixture of
N
a
O
H
and
N
a
2
C
O
3
contains x g of
N
a
O
H
and y g of
N
a
2
C
O
3
.The value of x to the nearest integer (in g) is
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Solution
Total mass=4g
Now,
m
o
l
e
s
o
f
N
a
O
H
=
m
o
l
e
s
o
f
N
a
2
C
O
3
=
a
W
N
a
O
H
+
W
N
a
2
C
O
3
=
4
g
⇒
40
a
+
106
a
=
4
⇒
40
a
+
106
a
=
4
⇒
a
=
4
146
m
o
l
∴
mass of NaOH
=
4
146
×
40
g
=
1.095
g
≈
1
g
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Standard XII Chemistry
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