A mixture has 18 g water and 414 g ethanol. The mole fraction of water in mixture is: (assume ideal behaviour of the mixture)

Molecular weight of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}=24+5+16+1=46$
Molecular mass of H₂O = 18
$$\begin{gathered} 414\mathrm{g}\text{ of }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\text{ has }\frac{414}{46}=9\text{ mole } \\ 18\mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}\text{ has }=\frac{18}{18}=1\mathrm{mole} \\ \text{ Mole fraction of water }=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{1}{1+9}=\frac{1}{10}=0.1 \end{gathered}$$