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

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Answer 1 · 2017-06-14T04:25:54

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

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.

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