benzene and toluene both have equal mole fraction in their mutual solution. what do you except about their mutual solution. what do you except about their mole fraction in vapour phase at same temperature
(given p°benzene =3p°toulene
)
Answers
Answer:
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