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

May 3, 2017

${\chi}_{\text{EtOH}} = 0.44 \ldots \ldots \ldots$

#### Explanation:

We have $44.4 \cdot m o l \cdot k {g}^{-} 1$ $E t O H \left(a q\right)$; i.e.

$\text{Concentration"="Moles of solute"/"Kilograms of solvent} = 44.4 \cdot m o l \cdot k {g}^{-} 1.$

We work from a starting point there is a solution prepared from $1.00 \cdot k g$ $\text{water}$, and thus there are $44.4 \cdot m o l$ $E t O H$:

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

$= \frac{44.4 \cdot m o l}{44.4 \cdot m o l + \left(\frac{1.0 \times {10}^{3} \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}\right)}$

$= \frac{44.4 \cdot m o l}{44.4 \cdot m o l + 55.5 \cdot m o l}$

${\chi}_{\text{EtOH}} = 0.44$

And for completeness, we calculate ${\chi}_{{H}_{2} O}$

$= \frac{55.5 \cdot m o l}{44.4 \cdot m o l + 55.5 \cdot m o l} = 0.55 .$

Of course, we knew that ${\chi}_{E} t O H + {\chi}_{\text{water}} = 1$ in a binary solution.