Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.

Assume that we have 100 g of solution (because percentage composition). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 x 2 + 6 x 1 + 16 x 2
                                     = 62 g mol^-1
$\begin{array}{l}\text{ Moles of }{\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2=\frac{20}{62}=0.322\mathrm{mol}\\ \text{ Moles of water }=\frac{80\mathrm{g}}{18{\mathrm{gmol}}^{-1}}=4.444\mathrm{mol}\end{array}$
$\begin{array}{l}{\mathrm{\chi }}_{\text{glycol }}=\frac{\text{ moles of }{\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2}{\text{ moles of }{\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2+\text{ moles of }{\mathrm{H}}_2\mathrm{O}}\\ =\frac{0.322\mathrm{mol}}{0.322\mathrm{mol}+4.444\mathrm{mol}}=0.068\end{array}$