Question

A solution is composed of a `40.0*g` mass of methanol, a `60.0*g` mass of ethanol. If the vapour pressures of pure methanol, and pure ethanol are `88.7*mm*Hg` and `44.5*mm*Hg` at this temperature, what is the vapour pressure of this solution?

Answer 1

The vapour pressure of the solution is proportional to mole fractions of the components of the solution....

Explanation:

The constant of proportionality is the mole fraction....

`chi_"component"="moles of component"/"total moles in solution"`

`chi_"MeOH"=((40.0*g)/(32.04*g*mol^-1))/((40.0*g)/(32.04*g*mol^-1)+(60.0*g)/(46.07*g*mol^-1))`

`chi_"EtOH"=((60.0*g)/(46.07*g*mol^-1))/((40.0*g)/(32.04*g*mol^-1)+(60.0*g)/(46.07*g*mol^-1))`

`"Vapour pressure of solution"=(chi_"MeOH"xx88.7+chi_"EtOH"xx44.5)*mm*Hg`.

Note that `chi_"MeOH"+chi_"EtOH"=1`, do you agree...?

The vapour of this solution, will be enriched with respect to the more volatile component...I leave you to do this calculation....