Composition of the original mixture 50% MeoAc (v/v) 50%, ROAc (v/v)%ur compound)

Question

Composition of the original mixture 50% MeoAc (v/v) 50%, ROAc (v/v)%ur compound) calculate (b) and (c) from ROAc (wt/wt) MeoAc (wt/wt) not from GC data c) NMeowe. NEATLY show calculations: 0 32 0.93 Mixture Composition of the distillate. NEATLY show all calculations. a) Using your new mole ratio above, calculate what you would expect the boiling point for your mixture to be (vapor pressure tables provided in lecture notes). Find the temperature closest to torr. composition of the vapors that first escape into the distillate (note, this is theoretically the highest purity you could obtain if we only saved the first drops of the simple distillate).

Explanation / Answer

Dear Candidate you have posted two questions, as per guidelines we solve here first question.

Q1.

MeOAc 50%(v/v)

PropOAc 50%(v/v)

it means 100mL of the mixture contains 50mL MeOAc and 50mL PropOAc

Mass of MeOAc = density X volume = 0.891 X 50 = 44.55g

Mass of MeOAc = density X volume = 0.932 X 50 = 46.6g

mass % of MeOAc = mass of MeOAc / total mass = 46.6 / (46.6 + 44.55) = 51.12 %(w/w)

mass % of PropOAc = mass of PropOAc / total mass = 44.55 / (46.6 + 44.55) = 48.875 %(w/w)

NMeOAC = mass of MeOAc / molar mass of MeOAc = 46.6 / 74.08 = 0.63

NPropOAC = mass of PropOAc / molar mass of PropOAc = 46.6 / 102.1 = 0.46

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