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Question
Chemistry
When 2.46 g of a hydrated salt (MSO4.xH2O) is completely dehydrated, 1.20 g of anhydrous salt is obtained. If the molecular weight of anhydrous salt is 120 g mol-1 what is the value of x ?
Q. When
2.46
g
of a hydrated salt (
MS
O
4
.
x
H
2
O
) is completely dehydrated,
1.20
g
of anhydrous salt is obtained. If the molecular weight of anhydrous salt is
120
g
m
o
l
−
1
what is the value of
x
?
2275
232
KEAM
KEAM 2016
Some Basic Concepts of Chemistry
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A
2
0%
B
4
6%
C
5
6%
D
6
6%
E
7
6%
Solution:
2.46
g
MS
O
4
⋅
x
H
2
O
⟶
Δ
1.20
g
MS
O
4
+
xg
H
2
O
Molecular weight of
MS
O
4
=
120
g
/
m
o
l
Molecular weight of
MS
O
4
⋅
x
H
2
O
=
120
g
+
x
×
18
g
(
120
g
+
x
.18
g
)
of
MS
O
4
⋅
x
H
2
O
on complete dehydration gives
120
g
of
MS
O
4
.
1
g
gives
=
120
+
18
⋅
x
120
Then,
2.46
g
of
MS
O
4
⋅
x
H
2
O
gives
120
+
18
x
120
×
2.4
g
which is equal to
1.20
g
∴
120
+
18
x
120
×
2.46
=
1.20
.
120
+
18
x
295.2
=
1.20
295.2
=
1.20
×
120
+
1.20
×
18
x
295.2
=
144
+
21.6
x
295.2
−
144
=
21.6
x
x
=
21.6
151.2
=
7
x
=
7
∴
The value of
x
=
7