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% ethylene glycol solution means that 20 g of ethylene glycol is present in 100 g of solution or 20 g of ethylene glycol is present is 80 g of water .
Molar mass of `C_(2)H_(6)O_(2) = 2 xx12+6 xx 1+2 xx 16 = 62 g mol^(-1)`
Moles of `C_(2)H_(6)O_(2) = (20 g)/(62 g mol^(-1)) = 0.322 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)/(0.322 + 4.444) = 0.068`
`x_("water") = (4.444)/(0.322+ 4.444) = 0.932`
or `" " x_("water") = 1 - 0.068 = 0.932`