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

Apr 27, 2017

Well, in aqueous solution, the solvent is water. By definition, $\text{4.4 molal}$ is $\text{4.4 mols solute/kg solvent}$. Therefore, you can assume that for $\text{4.4 mols ethanol}$, you have $\text{1 kg water}$.

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

$=$ $\text{55.51 mols water}$

Knowing the $\text{mol}$s of solute and of solvent...

$\textcolor{b l u e}{{\chi}_{E t O H}} = {n}_{E t O H} / \left({n}_{E t O H} + {n}_{{H}_{2} O}\right)$

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

$= \textcolor{b l u e}{0.0741}$

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 $\text{2.22 mols}$ of solute, thereby giving us $\text{0.5 kg}$ of water. We'd get the same result.