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

1 Answer
Nov 23, 2016

Answer:

#P_"solution"~=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"#

#chi_"MeOH"="Moles of MeOH"/"Moles 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)))#

#=# #(3.12*mol)/(3.12*mol+2.17*mol)=0.590#. #chi# is dimensionless; why?

We could do the same thing to get #chi_"EtOH"#, however this is a binary solution, and the mole fractions of each component must sum to #1#.

Thus #chi_"EtOH"=1-chi_"MeOH"=1-0.590=0.410#

And thus #P_"solution"=P_"MeOH"+P_"EtOH"#

#=# #chi_"MeOH"xx88.7*mm*Hg+chi_"EtOH"xx44.5*mm*Hg#

#=# #0.410xx88.7*mm*Hg+0.590xx44.5*mm*Hg#

#=# #??*mm*Hg#

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