Question

What is the mol fraction of ethanol in a `"4.4 mol/kg"` aqueous solution?

Answer 1

Well, in aqueous solution, the solvent is water. By definition, `"4.4 molal"` is `"4.4 mols solute/kg solvent"`. Therefore, you can assume that for `"4.4 mols ethanol"`, you have `"1 kg water"`.

`cancel"1 kg water" xx (1000 cancel"g")/cancel"1 kg" xx "1 mol water"/(18.015 cancel"g")`

`=` `"55.51 mols water"`

Knowing the `"mol"`s of solute and of solvent...

`color(blue)(chi_(EtOH)) = n_(EtOH)/(n_(EtOH) + n_(H_2O))`

`= "4.44 mols"/("4.44 mols" + "55.51 mols")`

`= color(blue)(0.0741)`

As note, it does not matter what temperature you are at. The molal is temperature-independent, because it is based on the mass of the solvent, not its volume.

You could even check what would happen if we had `"2.22 mols"` of solute, thereby giving us `"0.5 kg"` of water. We'd get the same result.