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A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K the
vapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. The
mixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off to
the second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.
vapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. The
mixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off to
the second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.
Imp : you should answer only the numeric value
Correct answer is '0.40'. Can you explain this answer?
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A binary liquid mixture of benzene and toluene contains 20 mol% of ben...
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A binary liquid mixture of benzene and toluene contains 20 mol% of ben...
's law.
a) What is the mole fraction of benzene in the vapour phase when the mixture is at its boiling point?
b) What is the total pressure of the mixture when it is at its boiling point?
c) What is the composition (in mole fraction) of the liquid phase when the mixture is at its boiling point?
Solution:
a) Since the mixture follows Raoult's law, the vapour pressure of the mixture at its boiling point is the sum of the vapour pressures of the pure components at that temperature. At the boiling point, the vapour pressure of the mixture is equal to the atmospheric pressure, which is 101.3 kPa. Therefore, we can write:
0.2 x 101.3 kPa (benzene) + 0.8 x 101.3 kPa (toluene) = 92 kPa (benzene) x + 35 kPa (toluene) (1 - x)
where x is the mole fraction of benzene in the vapour phase. Solving for x, we get:
x = 0.463
Therefore, the mole fraction of benzene in the vapour phase at the boiling point is 0.463.
b) The total pressure of the mixture at the boiling point is equal to the atmospheric pressure, which is 101.3 kPa.
c) At the boiling point, the composition (in mole fraction) of the liquid phase is given by:
x(benzene) = P(benzene) / P(total) = 0.2 x 101.3 kPa / 101.3 kPa = 0.2
x(toluene) = 1 - x(benzene) = 0.8
Therefore, the composition of the liquid phase at the boiling point is 0.2 mole fraction benzene and 0.8 mole fraction toluene.
a) What is the mole fraction of benzene in the vapour phase when the mixture is at its boiling point?
b) What is the total pressure of the mixture when it is at its boiling point?
c) What is the composition (in mole fraction) of the liquid phase when the mixture is at its boiling point?
Solution:
a) Since the mixture follows Raoult's law, the vapour pressure of the mixture at its boiling point is the sum of the vapour pressures of the pure components at that temperature. At the boiling point, the vapour pressure of the mixture is equal to the atmospheric pressure, which is 101.3 kPa. Therefore, we can write:
0.2 x 101.3 kPa (benzene) + 0.8 x 101.3 kPa (toluene) = 92 kPa (benzene) x + 35 kPa (toluene) (1 - x)
where x is the mole fraction of benzene in the vapour phase. Solving for x, we get:
x = 0.463
Therefore, the mole fraction of benzene in the vapour phase at the boiling point is 0.463.
b) The total pressure of the mixture at the boiling point is equal to the atmospheric pressure, which is 101.3 kPa.
c) At the boiling point, the composition (in mole fraction) of the liquid phase is given by:
x(benzene) = P(benzene) / P(total) = 0.2 x 101.3 kPa / 101.3 kPa = 0.2
x(toluene) = 1 - x(benzene) = 0.8
Therefore, the composition of the liquid phase at the boiling point is 0.2 mole fraction benzene and 0.8 mole fraction toluene.
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A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer?
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A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer?.
A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K thevapour pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off tothe second decimal place) of benzene in contact with this liquid mixture at 350 K is ________.Imp : you should answer only the numeric valueCorrect answer is '0.40'. Can you explain this answer?.
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