NaOH aqueous solution is labelled as 10% (w/v). Density of the solution is 1.02 g/ml. Then the mole fraction of the solute in the solution is

$\frac{\mathrm{W}}{\mathrm{V}}\%\text{ of aq. }\mathrm{NaOH}=10\%$

${\mathrm{W}}_{\mathrm{NaOH}}=10\mathrm{g}$

${\mathrm{V}}_{\text{sol. }}=100\mathrm{ml}$

${\mathrm{d}}_{\text{sol. }}=1.02 \mathrm{g}/\mathrm{ml}$

${\mathrm{W}}_{\text{sol. }}=100(1.02)=102\mathrm{g}$

${\mathrm{W}}_{{\mathrm{H}}_2\mathrm{O}}=(102-10)\mathrm{g}=92\mathrm{g}$

${\mathrm{X}}_{\mathrm{NaOH}}=\frac{{\mathrm{n}}_{\mathrm{NaOH}}}{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}+{\mathrm{n}}_{\mathrm{NaOH}}}=\frac{\left(\frac{10}{40}\right)}{\left(\frac{92}{18}\right)+\left(\frac{10}{40}\right)}$

$=\frac{0.25}{5.11+0.25}=\frac{0.25}{5.36}=0.0466$