Question

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

Answer 1

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.