Mole fraction of ethanol in ethanol water mixture is $0.25$ . Hence,the percentage concentration of ethanol by weight of mixture is:

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Solution

$x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = 0.25 (ethanol) (solute) (M w_{2} = 46 g)$
$x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = 0.75 (water) (solvent) (M w_{1} = 18 g)$
$\frac{x_{1}}{x_{2}} = \frac{n_{1}}{n_{2}} = \frac{3}{1} \Rightarrow \frac{W_{1} \times M w_{2}}{M w_{1} \times W_{2}} = \frac{3}{1}$
$\Rightarrow \frac{W_{1} \times 46}{18 \times W_{2}} = 3$
$\Rightarrow \frac{W_{1}}{W_{2}} = 1.17$
$\Rightarrow W_{1} = 1.17 W_{2}$
$∴ \frac{W_{2}}{W_{1} + W_{2}} = 0.46$ .

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