A mixture of benzene and toluene, the vapour pressure expression is P = [(180 x Xb) + 90] where Xb is the mole fraction of benzene in liquid phase. The mixture is prepared by mixing 780 grams of benzene and 736 grams of toluene. At 50 degree celsius, the vapour above the solution is condensed. Then condensed liquid further establishes equilibrium with the vapour phase. Calculate the mole fraction in the final vapour phase
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12th
Chemistry
Solutions
Vapour Pressure of Liquid Solutions
Vapour Pressure of a mixtur...
CHEMISTRY
Vapour Pressure of a mixture of benzene and toluene is given by P=179X
B
+92, Where X
B
is mole fraction of benzene.
If Vapours are removed and condensed in to liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate :
December 27, 2019avatar
Deepansha Gaikwad
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ANSWER
Given P=179X
B
+92
For pure C
6
H
6
,X
B
=1
∴P
B
o
=179+92=271mm
For pure C
7
H
8
,X
B
=0
∴P
T
o
=179×0+92=92mm
Now, P
M
=P
B
o
X
B
+P
T
o
X
T
=(271×
12+8
12
)+(92×
12+8
8
)
=199.4mm
as, Moles of C
6
H
6
=
78
936
=12
Moles of C
7
H
8
=
92
736
=8
Now, mole fraction of C
6
H
6
in vapour phase of initial mixture X
B
1
=
199.4
162.6
and that of toulene, X
T
1
=
199.4
36.8
∴
X
T
1
X
B
1
=
36.8
162.6
=4.418
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