To answer this questuon one should know the mass of ethanol in a sample of glacial acetic acid.
For example if m(C2H5OH)=0.05gm(C_2H_5OH) =0.05 gm(C2H5OH)=0.05g
Then w(C2H5OH)=0.05g1.78g=0.028w(C_2H_5OH) = \frac{0.05g}{1.78 g} = 0.028w(C2H5OH)=1.78g0.05g=0.028
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