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

1 Answer
Apr 27, 2017

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.