Answer
$X_{water}=0.677$
$X_{ethanol}=0.323$
Work Step by Step
Number of moles of $C_{2}H_{5}OH=\frac{55.0\,g}{46.07\,g/mol}$
$=1.1938\, mol$
Number of moles of water= $\frac{45.0\,g}{18.0\,g/mol}$
$=2.50\, mol$
Mole fraction of water,
$X_{water}=\frac{\text{moles of water}}{\text{moles of water}+\text{moles of ethanol}}$
$=\frac{2.50\,mol}{2.50\,mol+1.1938\,mol}$
$=0.677$
Mole fraction of ethanol,
$X_{ethanol}=\frac{\text{moles of ethanol}}{\text{moles of water}+\text{moles of ethanol}}$
$=\frac{1.1938\,mol}{2.50\,mol+1.1938\,mol}=0.323$
