Answer
$X_{ethanol}=0.274$
$X_{water}=0.726$
Work Step by Step
Mass of ethanol= $55.0\,mL\times0.789\,g/mL=43.395\,g$
Mass of water= $45.0\,mL\times1.00\,g/mL=45.0\,g$
Number of moles of $C_{2}H_{5}OH=\frac{43.395\,g}{46.07\,g/mol}$
$=0.9419\, mol$
Number of moles of water= $\frac{45.0\,g}{18.0\,g/mol}$
$=2.50\, mol$
Mole fraction of water,
$X_{water}=\frac{\text{moles of water}}{\text{moles of water}+\text{moles of ethanol}}$
$=\frac{2.50\,mol}{2.50\,mol+0.9419\,mol}$
$=0.726$
Mole fraction of ethanol,
$X_{ethanol}=\frac{\text{moles of ethanol}}{\text{moles of water}+\text{moles of ethanol}}$
$=\frac{0.9419\,mol}{2.50\,mol+0.9419\,mol}=0.274$