# Dichloromethane solubility in water

I heard dichloromethane is soluble in water 1:50, but I also heard it is not because it can't form hydrogen bonds. I'm confused. Is there a way to extract it? Would putting the water to $40~\mathrm{^\circ C}$ make it evaporate trough water?

• 1:50 is misleading. Wikipedia gives dichloromethane’s solubility in water as $20\,\mathrm{\frac{g}{l}}$ which is approximately 1:50 w/w, but ratios are usually understood v/v. (weight/weight versus volume/volume) – Jan Sep 30 '15 at 9:57
• @Jan That's a good start for an answer... – Martin - マーチン Sep 30 '15 at 11:11
• "it is not because it can't form hydrogen bonds" - Just bear in mind that solubility isn't a black and white thing. Dichloromethane is pretty insoluble in water (as opposed to something like ethanol) because it doesn't H-bond with water, but that doesn't mean it is entirely insoluble. – orthocresol Sep 30 '15 at 12:39

The solubility of dichloromethane in water (as well as the solubility of water in dichloromethane) depends on temperature. At a temperature of $T=20\ \mathrm{^\circ C}$, the solubility amounts to $17.6\ \mathrm g$ dichloromethane per $1\ \mathrm{kg}$ water; i.e. the mass fraction of dichloromethane is $w_{\ce{C2H2Cl2}}=0.0173$ or $w_{\ce{C2H2Cl2}}=1.73\,\%$. [source: “Aqueous Solubility and Henry’s Law Constants of Organic Compounds”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.]
You cannot simply evaporate pure dichloromethane (boiling point: $40\ \mathrm{^\circ C}$) from such mixtures since dichloromethane and water form an azeotrope, which boils at $38.8\ \mathrm{^\circ C}$. The mass fraction of water in this azeotrope is about $w_{\ce{H2O}}=1\,\%$.