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) = 12 xx 2 + 1 xx 6 + 16 xx 2 " 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_("glucol") = ("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") = (4.443 mol)/(0.032 mol + 4.444 mol ) = 0.932 `
Mole fraction of water can also be calculated as: ` 1 - 0.068 = 0.932 `