There are two separate phases, one of which is 12% phenol and the other is 60% phenol, from a 210 gram phenol-water mixture at a temperature of 60°C with a composition of 35% phenol. Calculate the masses of these phases.
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- Aniline, C6H5NH2, and hexane, C6H14, form partially miscible liquid–liquid mixtures at temperatures below 69.1 °C. When 42.8 g of aniline and 75.2 g of hexane are mixed together at a temperature of 67.5 °C, two separate liquid phases are formed, with mole fractions of aniline of 0.308 and 0.618. (i) Determine the overall mole fraction of aniline in the mixture. (ii) Use the lever rule to determine the relative amounts of the two phases.Phenol and water form non-ideal liquid mixtures. When 7.32 g of phenol and 7.95 g of water are mixed together at 60 °C they form two immiscible liquid phases with mole fractions of phenol of 0. 0 42 and 0 .161 . (a) Calculate the overall mole fraction of phenol in the mixture. (b) Use the lever rule to determine the relative amounts of the two phases .Phenol and water form non-ideal liquid mixtures. When 7.32 g of phenol and 7.95 g of water are mixed together at 60 °C they form two immiscible liquid phases with mole fractions of phenol of 0.042 and 0.161. (i) Calculate the overall mole fraction of phenol in the mixture. (ii) Use the lever rule to determine the relative amounts of the two phases.
- Aniline, C6H5NH2, and hexane, C6H14 , form partially miscible liquid - liquid mixtures at temperatures below 69. 1 °C. When 42.8 g of aniline and 75 .2 g of hexane are mixed together at a temperature of 67.5 °C, two separate liquid phases are formed, with mole fractions of aniline of 0.308 and 0.618.(a) Determine the overall mole fraction of aniline in the mixture. (b) Use the lever rule to determine the relative amounts of the two phases.At a temperature of 302.8 K, mixtures of phenol and water exist as two separate phases with mole fractions of phenol of 0.30 and 0.93. Use the lever rule to determine the relative amounts of the two phases for a mixture with an overall mole fraction of phenol of 0.45.The vapour pressure of pure liquid A at 300 K is 76.7 kPa and that of pure liquid B is 52.0 kPa. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of A in the vapour is 0.350. Calculate the total pressure of the vapour and the composition of the liquid mixture.
- Suppose that in a phase diagram, when the sample was prepared with the mole fraction of component A equal to 0.40 it was found that the compositions of the two phases in equilibrium corresponded to the mole fractions xA,α = 0.60 and xA,β = 0.20. What is the ratio of amounts of the two phases?There are two separate phases, one with a composition of 12% phenol and the other with a composition of 60% phenol, from a 210 gram phenol-water mixture with 35% phenol at 60oC. Calculate the masses of these phases.A mixture of phenol and water, under certain conditions of temperature and composition, forms two separate liquid phases, one rich in phenol and the other rich in water. At 30⁰ the compositions of the upper and lower layers are 70% and 9% by mass phenol, respectively. If 40kg of phenol and 60kg of water are mixed and the layers are allowed to separate at 30⁰C, what will be the weight of the two layers?
- A vapour solution with a mole percentage of 10% benzene is cooled gradually. At what temperature does liquid first appear?Calculate the Gibbs energy, entropy, and enthalpy of mixing when 0.50 mol C6H14 (hexane) is mixed with 2.00 mol C7H16 (heptane) at 298 K. Treat the solution as ideal.There are two separate phases, one with a composition of 12% phenol and the other with a composition of 60% phenol, from a 210 gram phenol-water mixture at 60 °C with a composition of 35% phenol. Calculate the masses of these phases.