Question

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?

Answer 1

`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.