Benzene and toluene from two ideal solution A and B at 313 K. Solution A (total pressure $P_{A}$ ) contains equal mole of toluene and benzene. Solution B contains equal masses of both A (total pressure $P_{B}$ ). The vapour pressure of benzene and toluene are 160 and 60 mm Hg respectively at 313 K. Calculate the value of $P_{A} + P_{B}$
(Round-off the answer).

Open in App

Solution

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 $P_{T} = P_{T}^{0} X_{T} = 60 m m H g \times 0.5 = 30 m m H g$ .
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 $P_{B} = P_{B}^{0} X_{B} = 160 m m H g \times 0.5 = 80 m m H g$ .
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 $P_{A} = P_{T} + P_{B} = 30 m m H g + 80 m m H g = 110 m m H g$ .
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 $P_{T} = P_{T}^{0} X_{T} = 60 m m H g \times 0.4588 = 27.53 m m H g$ .
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 $P_{B} = P_{B}^{0} X_{B} = 160 m m H g \times 0.5412 = 86.592 m m H g$ .
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 $P_{B} = P_{T} + P_{B} = 27.53 m m H g + 86.592 m m H g = 114.122 m m H g$ .
The value of the ratio $\frac{P_{A}}{P_{B}}$ is $\frac{P_{A}}{P_{B}} = \frac{110 m m H g}{114.122 m m H g} = 0.964$

Suggest Corrections

Similar questions

Benzene and toluene form nearly ideal solution .At 313K the vapour pressure of pure benzene is 150mm Hg and of pure toluene is 50 mm Hg .calculate vapour pressure of mixture of these two containing their equal masses at 313 K .

Q. Benzene and toluene form ideal solution occur the entire range of composition The vapour pressure of pure benzene and toluene at

300 K

are

50 : 71

H g

and

32.06

H g

respectively calculate the mole fraction of benzene is vapour phase if

80 g

of benzene is mixed [ 100 g of toluene ].

Q. A solution containing benzene and toluene in the mole ratio of 6:4 has a vapour pressure of 667.06 mm of Hg at

85^{o} C

.A solution at the same temperature containing benzene and toluene in the ratio of 4:6 has a vapour pressure of 559.74. Calculate the mole ratio of benzene-toluene which would boil at this temperature.Atmospheric pressure =760 mm of Hg.
if ratio is x:y then x+y is

Q. At

40^{o} C

, the vapour pressure of pure liquids, benzene and toluene, are 160 mm

H g

and 60 mm

H g

, respectively. At the same temperature, the vapor pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:

Q. An ideal solution of benzene and toluene boils at

1 a t m

pressure at

90^{\circ} C

. At

90^{\circ} C

benzene has a vapour pressure of 1022 mm of Hg and toluene has a vapour pressure of 406 mm of Hg. The mole fraction of benzene in the solution:

Join BYJU'S Learning Program