What is the total pressure exerted by a mixture of 0.45 kg mole of benzene, 0.44 kg mole of toluene and 0.23 kg mole of o-xylene at 100°C, if their vapor pressures at 100°C are 1340, 560 and 210 mmHg respectively ?
Correct answer: C. 801.5
- A. 756.2
- B. 780.5
- C. 801.5
- D. 880.5
Explanation
Using Raoult’s law, P = ΣxiPi, with total amount 1.12 kg mol gives mole fractions 0.4018, 0.3929 and 0.2054. Thus P = 0.4018(1340) + 0.3929(560) + 0.2054(210) ≈ 801.5 mmHg.
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