The solution The initial point is in the one$-$phase region. When heated it boils at $350K(a_2)$ giving a vapour of composition $x_B=0\mathrm{.}66(b_1).$ The liquid gets richer in $B,$ and the last drop $($of pure B$)$ evaporates at $390K.$ The boiling range of the liquid is therefore $350-390K.$ If the initial vapour is drawn off, it has a composition $x_B=0\mathrm{.}66.$ Cooling the distillate corresponds to moving down the $x_B=0\mathrm{.}66$ isopleth. At $330K,$ for instance, the liquid phase has composition $x_B=0\mathrm{.}87,$ the vapour $x_B=0\mathrm{.}49$; their relative proportions are $1$:$6.$ At $320K$ the sample consists of three phases: the vapour and two liquids. One liquid phase has composition $x_B=0\mathrm{.}30$; the other has composition $x_B=0\mathrm{.}80$ in the ratio $0\mathrm{.}52$:$1.$ Further cooling moves the system into the two$-$phase region, and at $298K$ the compositions are $0\mathrm{.}20$ and $0\mathrm{.}90$ in the ratio $0\mathrm{.}82$:$1.$ As further distillate boils over, the overall composition of the distillate becomes richer in $B.$ When the last drop has been condensed the phase composition is the same as at the beginning.