Chemistry, asked by asmazed6993, 8 months ago

Boiling point and Raoult’s law.for system benzene-toluene

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Answered by saxenaseema07054
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Explanation:

Boiling Point and Raoult's Law. For the system benzene-toluene, do as follows using the data in the following table 2. Vapor-Pressure and Equilibrium-Mole-Fraction Data for Benzene-Toluene System TABLE 11.1-1. Vapor Pressure Mole Fraction Benzene at 101.325 kPa Temperature Benzene Toluene °C kPa mm Hg kPa mm Hg 353.3 80.1 101.32 760 358.2 85 116.9 363.2 90 35.5 1016 54.0 405 368.2 95 155.7 1168 63.3 475 373.2 100 179.2 1344 74.3557 378.2 105 204.2 1532 86.0 383.8 110.6 240.0 1800 101.32 760 1.000 0.780 1.000 877 0.900 345 0.5810.632 46.0 0.411 0.258 0.130 0.456 0.261 645 0 a. At 378.2 K, calculate ya and xA using Raoult's Law. b. If a mixture has a composition of xA 0.40 and is at 358.2 K and 101.32 kPa pressure, will it boil? If not, at what temperature will it boil and what will the composition of the vapor first coming off

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