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Calculate the mole fraction of ethylene glycol (C_(2)H_(6)O_(2)) and water in a solution containing 20% of C_(2)H_(6)O_(2) by mass. |
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Answer» Solution :`20%` of `C_(2)H_(6)O_(2)` by mass MEANS 20 g of `C_(2)H_(6)O_(2)` are present in 100 g of the solution, i.e., `{:("Mass of SOLUTE "(C_(2)H_(6)O_(2))=26g,,,"Mass of solvent "(H_(2)O)=100-20=80g),("Molar mass of "C_(2)H_(6)O_(2)=62"g mol"^(-1),,,"Molar mass of"H_(2)O=18"g mol"^(-1)),(therefore" No, of moles of "C_(2)H_(6)O_(2)=(20)/(62)=0.322,,,"No. of moles of "H_(2)O=(80)/(18)=4.444):}` `"Mole FRACTION of "C_(2)H_(6)O_(2)" in the solution"=(.^(n)C_(2)H_(6)O_(2))/(.^(n)C_(2)H_(6)O_(2)+.^(n)H_(2)O)=(0.322)/(0.322+4.444)=0.068` `"Mole fraction of "H_(2)O" in the solution "=1-0.068=0.932` |
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