A 5.2 molal aqueous solution of methyl alcohol, CH₃OH is supplied. What is the mole fraction of methyl alcohol in the solution ?

$\begin{array}{l}\text{ Molality }=\frac{\text{ Moles of solute }}{\text{ Mass of solvent (in kg) }}\\ =\frac{5.2{\mathrm{molCH}}_3\mathrm{OH}}{1{\mathrm{kgH}}_2\mathrm{O}} [1\mathrm{kg}=1000\mathrm{g}]\end{array}$
${\mathrm{n}}_1\left({\mathrm{CH}}_3\mathrm{OH}\right)=5.2,{\mathrm{n}}_2\left({\mathrm{H}}_2\mathrm{O}\right)=\frac{1000}{18}=55.56$
$\therefore {\mathrm{n}}_1+{\mathrm{n}}_2=5.20+55.56=60.76\mathrm{mol}$
$\therefore {\mathrm{x}}_{{\mathrm{CH}}_3\mathrm{OH}}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{5.2}{60.76}=0.086$