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

$\text{w}+\text{W}=?$          $\text{d}=1.02$          $\text{d}=\frac{\text{m}}{\text{v}}$

$\text{m}=\text{d}\times \text{v}$

$=1.02\times 100$

$=102\text{g}$

$\text{w}+\text{W}=102$

$\text{W}=102-10=92\text{g}$

$x_{\text{solute}}=\frac{0.25}{0.25+5.11}$

$=0.0466$