Question

What is the mole fraction with respect to ethanol of an aqueous solution whose `"molality"=44.1*mol*kg^-1`?

Answer 1

`chi_"EtOH"=0.44.........`

Explanation:

We have `44.4*mol*kg^-1` `EtOH(aq)`; i.e.

`"Concentration"="Moles of solute"/"Kilograms of solvent"=44.4*mol*kg^-1.`

We work from a starting point there is a solution prepared from `1.00*kg` `"water"`, and thus there are `44.4*mol` `EtOH`:

`chi_"EtOH"="moles of EtOH"/("moles of EtOH + moles of water")`

`=(44.4*mol)/(44.4*mol+((1.0xx10^3*g)/(18.01*g*mol^-1)))`

`=(44.4*mol)/(44.4*mol+55.5*mol)`

`chi_"EtOH"=0.44`

And for completeness, we calculate `chi_(H_2O)`

`=(55.5*mol)/(44.4*mol+55.5*mol)=0.55.`

Of course, we knew that `chi_EtOH+chi_"water"=1` in a binary solution.