What are the mole fractions of #"water, ethanol, acetic acid,"# in a #25%:25%:50%# mixture by mass?

1 Answer
Jun 14, 2017

Answer:

We assume a mass of #100*g.#....and come up mole fractions of #chi_(H_2O)=0.503; chi_("EtOH")=0.196; chi_("HOAc")=0.301#.....

Explanation:

We assume a mass of #100*g.#....and work out the molar quantities on that basis........

#"Moles of water"=(25*g)/(18.01*g*mol^-1)=1.39*mol#

#"Moles of ethanol"=(25*g)/(46.07*g*mol^-1)=0.543*mol#

#"Moles of acetic acid"=(50*g)/(60.05*g*mol^-1)=0.833*mol#

And thus...........

#chi_(H_2O)=(1.39*mol)/((1.39+0.543+0.833)*mol)=0.503#

#chi_("EtOH")=(0.543*mol)/((1.39+0.543+0.833)*mol)=0.196#

#chi_("HOAc")=(1.39*mol)/((1.39+0.543+0.833)*mol)=0.301#

And we note that #chi_(H_2O)+chi_"EtOH"+chi_"HOAc"=1.00#, as is required for a #Sigman_i# where #n_i# are the individual molar quantities.