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Standard XII
Chemistry
Enthalpy
Question
The latent heat of vapourization of water at
100
o
C
is
540
c
a
l
.
g
−
1
. Calculate the entropy increase when one mole of water at
100
o
C
is evaporated.
26
c
a
l
.
K
−
1
m
o
l
−
1
1.45
c
a
l
.
K
−
1
m
o
l
−
1
367
c
a
l
.
K
−
1
m
o
l
−
1
1.82
c
a
l
.
K
−
1
m
o
l
−
1
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Solution
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Solution:- (A)
26
c
a
l
/
m
o
l
−
K
As we know that,
Δ
S
=
Δ
H
T
Δ
H
=
540
c
a
l
/
g
Mol. wt. of water
=
18
g
Δ
H
(
per mole
)
=
Δ
H
(
per gm
)
×
Mol. wt.
Δ
H
(
per mole
)
=
540
×
18
=
9720
c
a
l
/
m
o
l
Now, as we know that,
Δ
S
=
Δ
H
T
Given
T
=
100
℃
=
(
273
+
100
)
=
373
K
∴
Δ
S
=
9720
373
=
26
c
a
l
/
(
m
o
l
.
K
)
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