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

1 Answer
May 3, 2017

Answer:

#chi_"EtOH"=0.44.........#

Explanation:

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.