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Calculation 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.

Answer»

Solution :`20%` of `C_(2)H_(6)O_(2)` by MASS means that 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))=20G",Mass of solvent "(H_(2)O)=100-20g=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`
`therefore"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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