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1-butanol and chlorobenzene form a minimum boiling point azeotropic system. The mole fraction of 1-butanol in the liquid (x) and vapour (y) phases at 1.000 atm is given below for a variety of boiling temperatures
T/K 396.57 393.94 391.60 390.15 389.03 388.66 388.57 X 0.1065 0.17 0.2646 0.3687 0.5017 0.6091 0.7171
Y 0.2859 0.3691 0.4505 0.5138 0.5840 0.6409 0.7070
Pure chlorobenzene boils at 404.86 K. (a) Construct the chlorobenzene-rich portion of the phase diagram from the data. (b) Estimate the temperature at which a solution whose mole fraction of 1-butanol is 0.300 begins to boil. (c) State the compositions and relative proportions of the two phases present after a solution initially 0.300 1-butanol is heated to 393.94 K

Sagot :

So a) so atm is 1.000 b) solution with mole fraction is 0.300 starts to boil, c) At 393.94K, x = 0.1700 and y = 0.3691.

(a) The mole fraction of 1-butanol in the liquid (x) and vapor (y) phases at 1.00 atm at various boiling temperatures is: the calculation is shown below.

Necessary: Now we must make a graph of x (or) y versus T/K. I don't have any tools with which to create this graph. So that we may compute the boiling point of a solution with a mole fraction of 1-butanol of 0.300 at 391 degrees Celsius, I can supply a rough graph (scanned).

(b) At 391.0 K, the smooth curves x and 7 cross 0.3 (mole fraction of 1-butanol), showing that the boiling point of the solution with a mole fraction of 0.300 starts to boil.

c) The compositions and relative amounts of the two phases (c) The proportions of either the two phases are mostly in inverse proportion to the distances of the mole fractions first from the composition point in question according to Liver's rule.

That is equal to =0.5315.

Based on the information provided: At 393.94K, x = 0.1700 and y = 0.3691.

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