If the overall mixture is 80% by mole methanol, will the mixture exist in one phase or two phases in equilibrium at 315 K? If it does phase split, what is the composition of the equilibrium liquid phases?
Consider the methanol + cyclohexane system in the figure.
A. The mixture exists in one liquid phase.
B. The mixture exists in two liquid phases. The composition of the “methanol rich” phase is 80% methanol, while the composition of the “cyclohexane rich” phase is 15% methanol.
C. The mixture exists in two liquid phases. The composition of the “methanol rich” phase is 90% methanol, while the composition of the “cyclohexane rich” phase is 70% methanol
D. The mixture exists in two liquid phases. The composition of the “methanol rich” phase is 70% methanol, while the composition of the “cyclohexane rich” phase is 26% methanol
E. The mixture exists in one vapor phase.
A. Correct. The point at the intersection of 0.8 mole fraction methanol and 315 K is outside the dome, so only one liquid phase exists.
B. Incorrect. You are misreading the graph. The mixture is at 315 K, not 305 K.
C. Incorrect. You are misreading the graph. Review Example 13-2 on how to read these graphs.
D. Incorrect. If you are outside the LLE dome, you do not read compositions on the curve because you are not in an LLE region.
E. Incorrect. Outside the LLE dome, a single liquid phase exists (not a vapor phase). At higher temperatures, however, a vapor phase will exist.
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