A solution has 1 : 4 $$mole$$  ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $${20^ \circ }C$$  are $$440\,mm$$  of $$Hg$$  for pentane and $$120$$ $$mm$$  of $$Hg$$  for hexane. The mole fraction of pentane in the vapour phase would be

Solution :
Total vapour pressure of mixture
= Vapour pressure of pentane in mixture + vapour pressure of hexane in mixture
Since, the ratio of pentane to hexane = 1 : 4
$$\therefore $$  $$Mole$$  fraction of pentane $$ = \frac{1}{5}$$
$$Mole$$  fraction of hexane $$ = \frac{4}{5}$$
= ( $$mole$$  fraction of pentane × vapour pressure of pentane ) + ( $$mole$$  fraction of hexane × vapour pressure of hexane)
$$\eqalign{ & = \left( {\frac{1}{5} \times 440 + \frac{4}{5} \times 120} \right) \cr & = 184\,mm \cr} $$
$$\because $$   Vapour pressure of pentane in mixture
= Vapour pressure of mixture × $$mole$$  fraction of pentane in vapour phase
$$88 = 184 \times mole$$    fraction of pentane in vapour phase
$$\therefore $$  $$Mole$$  fraction of pentane in vapour phase
$$ = \frac{{88}}{{184}} = 0.478$$