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If I have a mixture of 25 vol% methanol in water and I mix it with an equal volume of dichloromethane (which is miscible with methanol but not water) can I roughly say that both phases (i.e. water and dichloromethane) will have the same methanol concentration upon reaching equilibrium? Or will it vary significantly based on the activity of each mixture?

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    $\begingroup$ I doubt it. I would expect the MeOH to go into the water phase $\endgroup$ – Waylander Apr 8 '19 at 13:40
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    $\begingroup$ When you have two phases they is always some of each component in both phases. // (solubility of dichloromethane in pure water is 17.5 g/L @ 25 °C) // I'd expect the dichloromethane to contain both water and methanol, and the methanol/water phase will contain a small amount of dichloromethane. In certain proportions I'd expect one phase. $\endgroup$ – MaxW Apr 8 '19 at 15:38
  • $\begingroup$ What is needed is the ternary phase diagram for the methanol/water/dichloromethane system. // I looked and couldn't find one. But there must have been studies on this. $\endgroup$ – MaxW Apr 8 '19 at 15:45
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This is a partial phase diagram. That seems to indicate that systems with roughly 50% methanol are miscible.

Source: http://shodhganga.inflibnet.ac.in/bitstream/10603/9995/11/11_chapter%206.pdf

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