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 20g 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)=(20g)/(62g mol^(-1))=0.322` mol
Moles of water `= (80g)/(18g 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 mol + 4.444 mol)=0.932`
Mole fraction of water can also be calculated as : `1-0.068=0.932`