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toluene: $\ce{C6H5-CH3}$

cyclohexane: $\ce{C6H12}$


I think that answer would be:

  1. Toluene and cyclohexane do not dissolve in water (since they are nonpolar), so I would first separate the water from the mixture with a separator funnel.

  2. A mixture of cyclohexane and toluene would be separated by fractional distillation. Cyclohexane, which has a lower boiling point than toluene, would be eliminated from the mixture.

Is there anything wrong in my answer? Would there also be anything to add?

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    $\begingroup$ Sounds good to me. $\endgroup$ – Ivan Neretin Apr 12 at 15:43
  • $\begingroup$ Cyclohexane and toluene may probably be difficult to separate by distillation. Toluene is not very different from benzene. And i know that benzene and cyclohexane are forming an azeotropic mixture (59% benzene) which boils at 77°C, that is lower than the boiling points of benzene and of cyclohexane. So I would not be surprised to learn that toluene and cyclohexane are making an azeotropic mixture. This should be checked. $\endgroup$ – Maurice Apr 12 at 16:19
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    $\begingroup$ Cyclohexane and toluene would not make azeotrope. Fractional distillation would do the trick. $\endgroup$ – Mathew Mahindaratne Apr 12 at 16:21
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Distillation

The most effective and easiest way to separate the mixture of toluene and cyclohexane is simple distillation. Cyclohexane has boiling point is about 81°C, and toluene’s boiling point is 111°C.

References:

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    $\begingroup$ Simple distillation is not the most effective but easiest. If you look at the NMR of the first distillate, it is 1:2 mixture of toluene to cyclohexane. The residue is 8:5 mixture of toluene to cyclohexane. I made my point. $\endgroup$ – Mathew Mahindaratne Apr 12 at 16:43

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