At `298K` , the vapour pressure of pure liquid n-butane is 1823 torr and vapour pressure of pure n-pentane is 521 torr and form nearly an ideal solution.
a. Find the total vapour pressure at `298 K` of a liquid solution containing `10%` n-butane and `90%` n-pentane by weight,
b. Find the mole fraction of n-butane in solution exerting a total vapour pressure of 760 torr.
c. What is composition of vapours of two components (mole fraction in vapour state)?

Explanation

`P^@_("n-butane")=1823` torr, `P^@_("n-pentane")=521` torr
a. Given `10%` n-butane by weight = `10 g`
`90%` n-pentane by weight = `90 g`
`{:( :. ,"moles of n-butane"=10/58),(,"moles of n-pentane"=90/72):}({:("Mw of n-butane"=58),("Mw of n-pentane"=72):})`
`:. chi_("n-butane") = (10//58)/(10//58 + 90//72) = 0.122`
`chi_("n-pentane") =1-chi_("n-pentane") = 1-0.122 =0.878`
From Raoult's law,
`P_(t) =P_(A)@chi_(A) + P_(B)@chi_(A) = 0.122 xx 1823 + 0.878 xx 521`
= `679.84` torr
b. Given `P_("total") = 760 "torr", P@_("n-butane") =1823` torr.
`P_("n-pentane")` = 521 torr
Let mole fraction of n-butane =`y`
mole fraction n-pentane =`1-y`
Using Raoult's law, we have
`P_(t)=P_A^@chi_(A) + P_B^@chi_(B)`
`760 = y(1823) + (1-y)521`
`:. y=0.183`
c. The mole fraction in vapour state: `chi_("n-butane")^(V)=(P_("n-butane"))/(P_("total"))=(P^@_("n-butane") xx chi_("n-butane"))/P_("total")`
`= 0.122 xx 1823/679.84 =0.327`
`chi^V` is also called as vapour fraction.