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A solution is 25% ethanol ans 50% acetic acid by mass. Calculate the mole fraction of ethanol and acetic acid in the solution. |
Answer» Let us start with 100 g of the solutioon in which Mass of water =25g Mass of ethanol=25g Mass of acetic acid = 50g `n_(H_(2)O)=((25g))/((18gmol^(-1)))=1.388mol` `n_(C_(2)H_(5)OH)=((25g))/((46gmol^(-1)))=0.543 mol` `n_(CH_(3)COOH)=((50g))/((60gmol^(-1)))=0.833mol` Total number of moles =(1,388+0.543+0.833)=2.764 mole `x_(C_(2)COOH)" "=n_(C_(2)H_(5)OH)=((0.543 mol))/((2.764 mol))=0.196` `x_(CH_(3)COOH)" "n_(CH_(3)COOH)=((0.833 mol))/((2.764 mol))=0.301` |
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