Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25. Hence percentage of ethyl alcohol by weight is

If ethanol is 0.25, then ${\mathrm{H}}_2\mathrm{O}$ will be 0.75.

Mol wt of ethanol is 46.06 g/mol,

So, $0.25 mol=11.515g$ of ethanol

Mol wt of ${\mathrm{H}}_2\mathrm{O}$ is 18.01528g/mol,

So, $0.75 mol=13.51g$ of ${\mathrm{H}}_2\mathrm{O}$

Total $wt=25.02646 g$

wt % of etanol is $=\frac{11.515}{25.02646}\times 100=46.011\approx 46\mathrm{\%}$

$\frac{0.25\times 46}{0.25\times 46+0.75\times 18}=0.46$

$=46\%$