At a given temperature the vapour pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the mole fraction of benzene in the vapour phase over a solution of benzene and toluene with Xbenzene = 0.600.
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Explanation:
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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