Calculate the mole fraction of ethylene glycol `(C_(2)H_(6)O_(2))`
in a solution containing `20%` of `C_(2)H_(6)O_(2)` by mass.

Explanation

`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`