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 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g 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^(-1))=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)=0.068`
Similarly, `x_("water")=(4.444 mol)/(0.322+4.444 mol)=0.932`
Mole fraction of water an also be calculated as : `1-0.068=0.932`