The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Explanation

It is given that :
`P_(A)^(0) = 450 ` mm of Hg
`P_(B)^(0)` = 700 mm of Hg
`P_("Total") = 600 nm ` of Hg
From Roult.s law , we have
`P_("total") = P_(A) + P_(B)`
`rArr " " P_("total") = p_(A)^(0) x_(A) + p_(B)^(0) (1-x_(A))`
`rArr " " P_("Total") = p_(A)^(0) x_(A) + p_(B)^(0) - p_(B)^(0)x_(A)`
`rArr " " P_("Total") = (p_(A)^(0) - p_(B)^(0)) x_(A) + p_(B_^(0)`
`rArr" " 600 = (450-700) x_(A) + 700`
`rArr " " - 100 = - 250 x _(A)`
`rArr " " X_(A) = 0.4`
Therefore `" " x_(B) = 1 - x_(A)`
` = 1-0.4`
` = 0.6`
Now , `" " P_(A) = P_(A)^(0) x_(A)`
` = 450 xx 0.4`
= 180 mm of Hg
and ` " " P_(B) = P_(B)^(0) x_(B)`
` " " = 700 xx 0.6`
` " " = 420 ` mm of Hg.
Now , in the vapour phase : Mole fraction of liquid `A = P_(A) //(P_(A) + P_(B))`
` = 180 //(180+ 420)`
` = 180//600`
` = 0.30`
And, mole fraction of liquid B = 1-0 .30 = 0.70