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

Assume that we have 100g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20g of ethylene glycol and 80g of water.
Molar mass of `C_(2)H_(6)O_(2)=12xx2+1xx6+16xx2=62" g mol"^(-1)`
Moles of `C_(2)H_(6)O_(2)=(20"g")/(62" g mol"^(-))=0.322" mol"`
Moles of water `=(80"g")/(18" g mol"^(-1))=4.444" mol"`
`X_("glycol")=("moles of "C_(2)H_(6)O_(2))/("moles of "C_(2)H_(6)O_(2)+"moles of H"_(2)O)`
`=(0.322" mol")/(0.322" mol"+4.444" mol")=0.932`
Mole fraction of water can be calculated as : `1-0.068=0.932`