A solution, that we may consider to be ideal, is composed of a 100*g100⋅g masses EACH of ethanol and methanol....what is the vapour pressure of the solution?

A solution, that we may consider to be ideal, is composed of a 100*g100⋅g masses EACH of ethanol and methanol. If the vapour pressures of each PURE solvent at a given temperature are 44.5*mm*Hg44.5⋅mm⋅Hg and 88.7*mm*Hg88.7⋅mm⋅Hg respectively, what is the vapour pressure of each component, and the vapour pressure of the solution?

1 Answer
anor277
Nov 23, 2016

P_"solution"~=60*mm*HgPsolution≅60⋅mm⋅Hg

Explanation:

The partial pressure of each component in solution is proportional to its mole fraction.

Thus P_"MeOH"=chi_"MeOH"xxP_"pure MeOH"PMeOH=χMeOH×Ppure MeOH

chi_"MeOH"="Moles of MeOH"/"Moles of MeOH + Moles of EtOH"χMeOH=Moles of MeOHMoles of MeOH + Moles of EtOH

== ((100*g)/(32.04*g*mol^-1))/(((100*g)/(32.04*g*mol^-1))+((100*g)/(46.07*g*mol^-1)))100⋅g32.04⋅g⋅mol−1(100⋅g32.04⋅g⋅mol−1)+(100⋅g46.07⋅g⋅mol−1)

== (3.12*mol)/(3.12*mol+2.17*mol)=0.5903.12⋅mol3.12⋅mol+2.17⋅mol=0.590. chiχ is dimensionless; why?

We could do the same thing to get chi_"EtOH"χEtOH, however this is a binary solution, and the mole fractions of each component must sum to 11.

Thus chi_"EtOH"=1-chi_"MeOH"=1-0.590=0.410χEtOH=1−χMeOH=1−0.590=0.410

And thus P_"solution"=P_"MeOH"+P_"EtOH"Psolution=PMeOH+PEtOH

== chi_"MeOH"xx88.7*mm*Hg+chi_"EtOH"xx44.5*mm*HgχMeOH×88.7⋅mm⋅Hg+χEtOH×44.5⋅mm⋅Hg

== 0.410xx88.7*mm*Hg+0.590xx44.5*mm*Hg0.410×88.7⋅mm⋅Hg+0.590×44.5⋅mm⋅Hg

== ??*mm*Hg??⋅mm⋅Hg

As is typical in these solutions, the vapour is enriched with respect to the more volatile component, here methyl alcohol.

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