Mole fraction of ethanol-water mixture is 0.25. What will be the weight fraction of solute?

X₂ is the mole fraction of ethanol (solute), X₁ is the mole fraction of solvent (water), n₂ is the number of moles of solute and n₁ is the number of moles of solvent.

${\mathrm{\chi }}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.25\text{....(1)}$

${\mathrm{\chi }}_1=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=0.75\text{....(2)}$

Dividing (2) by (1)

$\begin{array}{l}\frac{{\mathrm{X}}_1}{{\mathrm{X}}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{3}{1}\\ \frac{{\mathrm{W}}_1\times {\mathrm{Mw}}_2}{{\mathrm{Mw}}_1\times {\mathrm{W}}_2}=\frac{3}{1}\\ \frac{{\mathrm{W}}_1\times 46}{18\times {\mathrm{W}}_2}=3\\ \frac{{\mathrm{W}}_1}{{\mathrm{W}}_2}=1.17\end{array}$

$\therefore \frac{{\mathrm{W}}_2}{{\mathrm{W}}_1+{\mathrm{W}}_2}=0.46$