benzene and toluene form nearly ideal solution .at 313k the vapour pressure of pure benzene is 160mm hg and of pure toluene is 60 mm hg .calculate vapour pressure of mixture of these two containing equal molecules of benzene and toluene and 1mole of benzene and 4mole of toluene
Answers
Answer:
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Explanation:
For solution A, the mole fractions of toluene and benzene are equal which comes out to be 0.5 each.
The partial pressure of toluene in the vapor phase is P
T
=P
T
0
X
T
=60 mm Hg×0.5=30mmHg.
Here, P
T
0
is the vapor pressure of pure toluene and X
T
is the mole fraction of toluene in liquid phase.
The partial pressure of benzene in the vapor phase is P
B
=P
B
0
X
B
=160 mm Hg×0.5=80mm Hg.
Here, P
B
0
is the vapor pressure of pure benzene and X
B
is the mole fraction of benzene in liquid phase.
The total pressure is P
A
=P
T
+P
B
=30 mm Hg+80 mm Hg=110 mm Hg.
For solution B the masses of toluene and benzene are equal .
The molar masses of benzene and toluene are 78.11 g/mol and 92.14 g/mol respectively.
Let 78.11 g of benzene and 78.11 g of toluene are present in the mixture.
This corresponds to 1 mole of benzene and 0.847 mole of toluene.
The mole fractions of benzene and toluene are 0.5412 and 0.4588 respectively.
The partial pressure of toluene in the vapor phase is P
T
=P
T
0
X
T
=60mmHg×0.4588=27.53mmHg.
Here, P
T
0
is the vapor pressure of pure toluene and X
T
is the mole fraction of toluene in liquid phase.
The partial pressure of benzene in the vapor phase is P
B
=P
B
0
X
B
=160mmHg×0.5412=86.592mmHg.
Here, P
B
0
is the vapor pressure of pure benzene and X
B
is the mole fraction of benzene in liquid phase.
The total pressure is P
B
=P
T
+P
B
=27.53mmHg+86.592 mm Hg=114.122 mm Hg.
The value of the ratio
P
B
P
A
is
P
B
P
A
=
114.122 mm Hg
110 mm Hg
=0.964