Question

Vapour pressure of `"C"_(6)"H"_(6)" and""C"_(7)"H"_(8)` mixture at `50^(@)"C"` is given by P(mm Hg)`=180X_("B")+90`, where `"X"_("B")` is the mole fraction of `"C"_(6)"H"_(6).` A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapour over this solution ar removed and condensed into liquid and again brought to the temperature of `50^(@)"C"`, what would be the new mole fraction of `"C"_(6)"H"_(6)` in the vapour state ?

Answer 1

Correct Answer - 0.93
`"C"_(6)"H"_(6)-"B,"" C"_(6)"H"_(7)-"T"`
`"n"_("B")=936/78," n"_("T")=736/92=8`
`"X"_("B")=("n"_("B"))/("n"_("B")+"n"_("T"))=12/(12+8)=0.6`
Now `"P"=("P"_("B")^(0)-"P"_("T")^(0))"X"_("B")+"P"_("T")^(0)`
`"P"_("A")^(0)=90"mm"," P"_("T")^(0)=270"mm"`
`"P"_("T")=0.6xx270," P"=180xx0.6+90`
`=162"mm "=198"mm"`
`"Y"_("B")=("P"_("B"))/"P"=162/198=9/11`
Mole fraction of benzene in vapour
Now on condensation
`"X"_("B")="Y"_("B")" X"_("B")=9/11`
`"P"_("B")=9/11xx270"mm"," P"=180xx9/11+90=90(29/11)"mm"`
`"X"_("B")="P"_("B")/"P"=(9/11xx270)/(90xx29/11)=27/29=0.93`