How do you calculate the χacetone and χcyclohexane in the vapor above the solution. P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr?

A solution is prepared by mixing 4.40 mol of acetone with 1.45 mol of cyclohexane at 30°C.

1 Answer

Answer:

Well by definition....the vapour pressure of EACH component is PROPORTIONAL to #chi_("component"(g))#...

#P_"component" = chi_("component"(g))P_"solution"#

Explanation:

Well in solution, the #"mole fraction"# is

#chi_("component"(l))="moles of component"/"Total moles in solution"#

So,

#chi_("acetone"(l))="moles of acetone"/"moles of acetone + moles of cyclohexane"#...

#=(4.40*mol)/(4.40*mol+1.45*mol)=0.752.#

And clearly....#chi_("cyclohexane"(l))=1-chi_("acetone"(l))=1-0.752=0.248#.

And so

#P_"acetone"=chi_("acetone"(l))xx229.5*mm*Hg=0.752xx229.5*mm*Hg=172.6*mm*Hg#

#P_"cyclohexane"=chi_("cyclohexane"(l))xx97.6*mm*Hg=0.248xx97.6*mm*Hg=24.2*mm*Hg#

#P_"solution"=P_"acetone"+P_"cyclohexane"={172.6+24.2}*mm*Hg#

#-=196.8*mm*Hg#

But in the vapour phase, the vapour pressure of EACH component is PROPORTIONAL to #chi_("component"(g))#...

#P_"component" = chi_("component"(g))P_"solution"#

As always, the vapour pressure is ENRICHED with respect to the more volatile component....here acetone...