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A solution contains 25% water, 25% ethanol and 505 aceticacid by mass. Calculate the mole fraction of each component. |
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Answer» SOLUTION :Let us suppose that the total mass of the solution is 100 g. Then AMOUNT of water = 25 g , `""` Amount of ethanol = 25 g `"25 g of water "=("25 g ")/("18 g MOL"^(-1))="1.388 moles"[because"Malar mass of water "="18 g mol"^(-1)]` `"50 g of ethanol "=("25 g")/("46 g mol"^(-1))="0.543 moles"[because "Molar mass of ethanol "(C_(2)H_(5)OH)="46 g mol"^(-1)]` `"50 g of ACETIC acid "=("50 g")/("60 g mol"^(-1))="0.833 moles"` `""[because"Molar mass of acetic acid "(CH_(3)COOH)="60 g mol"^(-1)]` `therefore"Mole fraction of water "=(1.388)/(1.388+0.543+0.833)=(1.388)/(2.764)=0.503` Mole fraction of ethanol `=(0.543)/(2.764)=0.196"and mole fraction of acetic acid "=(0.833)/(2.764)=0.301.` |
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