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Byju's Answer
Standard XII
Chemistry
System, Surroundings, Intensive, Extensive Properties
The entropy o...
Question
The entropy of vaporization of benzene is
85
J
K
−
1
m
o
l
−
1
. If
117
g benzene vaporizes at its normal boiling point, calculate the entropy change of the surrounding.
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Solution
Entropy of vapourization of benzene
Δ
s
=
85
J
k
−
1
m
o
l
−
1
Amunt of benzene
=
117
g
Molar mass of benzene
(
C
6
H
6
)
=
78
g
number of mole
=
117
78
=
1.5
m
o
l
e
Entropy charge of surrounding
=
−
85
×
1.5
=
−
127.5
J
K
−
1
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0
Similar questions
Q.
The entropy of vaporization of benzene is
85
J
K
−
1
m
o
l
−
1
. When
117
benzene vaporizes at its normal boiling point, the entropy change of surrounding is:
Q.
The entropy of vaporization of benzene is
85
J
K
−
1
m
o
l
−
1
.
When
117
g
benzene vapourization:
Q.
Calculate the boiling point of the liquid if its entropy of vaporization is
110
J
K
−
1
m
o
l
−
1
and the enthalpy of vaporization is
40.85
k
J
m
o
l
−
1
.
Q.
Ethanol boils at
78.4
o
C
and standard enthalpy of vaporization of ethanol is
42.4
k
J
m
o
l
−
1
. Calculate the entropy of vaporization of ethanol.
Q.
The entropy of vaporisation of a liquid is
58
J
K
−
1
m
o
l
−
1
. If
100
g
of its vapor condenses at its boiling point of
123
∘
C
, the value of entropy change for the process is
(Molar mass of the liquid
=
58
g mol
−
1
)
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