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.