Answer to Question #99907 in Physical Chemistry for Deema

Question

a) Sketch the temperature-composition ("T-x") phase diagram for acetone ((CH3)2CO) and chloroform (CHCl3) mixtures based on the following data:
- Boiling point of pure acetone: 56.2 oC
- Boiling point of pure chloroform: 61.2 oC
- Boiling point of the acetone-chloroform azeotrope: 64.7 oC
- Composition of the azeotrope (by weight): 21% acetone, 79% chloroform
- Atomic weights (in g/mol): H: 1.01, C: 12.01, O: 16.00, Cl: 35.45
Hint: First calculate the acetone mole fraction that corresponds to 21% acetone by weight.
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b) On your sketch of the phase diagram, explain how you would determine the composition of the vapor phase and the liquid phase for a system at 61.2 oC having an overall acetone mole fraction of z = 0.80

Answer 1

Solution.

1.

$M((CH3)2CO) = 58.09 \ g/mol$

$M(CHCl3) = 119.37 \ g/mol$

$n((CH3)2CO) = \frac{21 \ g}{58.09 \ \frac{g}{mol}} = 0.36 \ mol$

$n(CHCl3) = \frac{79 \ g}{119.37 \ \frac{g}{mol}} = 0.66 \ mol$

$N((CH3)2CO) = \frac{0.36}{0.66+0.36} \times 100 \% = 35.0 \%$

$N(CHCl3) = \frac{0.66}{0.66 + 0.36} \times 100 \%= 65.0 \%$

2.

In this sketch of the phase diagram, additional data on positive and negative deviations from the ideality of a non-ideal solution are needed. Therefore, it is necessary to draw a horizontal line at a temperature of 61.2 degrees and the intersection points will be equal to the molar fraction of the steam and liquid phases. The point on the left is the liquid phase and the point on the right is the vapor phase.

Answer:

2.

In this sketch of the phase diagram, additional data on positive and negative deviations from the ideality of a non-ideal solution are needed. Therefore, it is necessary to draw a horizontal line at a temperature of 61.2 degrees and the intersection points will be equal to the molar fraction of the steam and liquid phases. The point on the left is the liquid phase and the point on the right is the vapor phase.