What is the mole fraction with respect to ethanol of an aqueous solution whose $"molality"=44.1molkg^-1$ ?

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Answer 1 · 2017-05-03T17:08:27

$chi_"EtOH"=0.44.........$

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.

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