If you mix 20 moles of propyl acetate with 20 moles of a 75%/25% (by mol) mixture of water/acetic acid, what is the amount (by mole) and composition of the resulting equilibrium phases? Also, identify which compound would be considered the solute in this system and why.
Consider the ternary diagram below with compositions given as mole fractions.
First calculate the new composition of the mixture:
The figure below shows how the graph should be read to obtain the two resulting equilibrium phases.
The arrows indicate the direction to read for each compound. The red dot is the composition of the given mixture before it separates. The two green dots are the compositions of the two equilibrium phases of the given mixture. The resulting compositions of the two equilibrium phases are:
Note that these compositions are approximate. Final answers will vary, but should be similar to these values.
To find the amount of each phase, use the fact that there is essentially no propyl acetate in phase 2; all 20 moles are in phase 1.
The solute in this mixture is acetic acid (2) because it is present in both water (1) and propyl acetate (3) which act as the solvents.
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