Vapour-Liquid Equilibrium I have been studying flash distillation, and I am reading the VLE tables my textbook has provided. 
I am looking at these tables and I am not understanding the data, I think. 

Why is the vapour mole percent ($y_{benzene}$) and liquid mole percent of benzene ($x_{benzene}$) zero at 110.62 $^0 C$? Does that mean benzene just disappear from the flash distiller at that temperature?
Absolutely not. It means that for the mixture to be boiling at 110°C, it must not contain any benzene.