Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(@)C` is given by `P(mm Hg) = 179 X_(B) +92`, where `X_(R)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of `50^(@)C`. what would be mole fraction of `C_(6)H_(6)` in the vapour state?

Answer

Given, `P_(M)=179X_(B)+92`
`X_(B)` i.e., mole fraction of benzenen `=1`
`therefore P_(B)^(@)=179+92=271mm`
For pure `C_(7)H_(8);X_(B)=0`
`therefore P_(T)^(@)=179xx0+92=92mm`
(b)(936/78)/(936/78+(736)/(92))=(12)/(12+8)=0.6`
`P_(M)=179xx0.6+92=107.4+92`
`=199.4mm`
(c) Mole fraction of `C_(6)H_(6)` in vapour phase of initial mixture
`X_(L)'=P_(B)'//P_(M)`
`X_(B)'=[271xx0.6]//199.4=0.815`
Similarly, `X_(T)'=[92xx0.4]//199.4=0.185`
These fraction of vapours are taken out and condensed into. The liquid is again brought is again to `50^(@)C` to get again vapour-liquid equilibrium.
Mole fraction of components in vapour phase of initial `=` Molre fraction of components in liquid phase of II mixture
`therefore P_(M)=P_(B^(0).X_(B)'+P_(1)'.X_(T)'`
`=271xx0.815+92xx0.185mm`
New mole fraction of `C_(6)H_(6)` in vapour phase
`=P_(B)'//P_(M)=220.865//237.885=0.928`
New mole fraction of toluenen in vapour phase
`=1-0.928=0.072`