A mixture of ethanol $C_2H_6O$ and water has a mole fraction $0.235$ for alcohol and $0.765$ for water.
Calculate the percentage of alcohol in the mixture!
$1\; mol$ of the mixture $Q$ contains $0,235\; mol$ alcohol and $0,765\; mol$ water:
$m_{C_2H_6O}=0.235\cdot M_{C_2H_6O}=10.8\;g$
$m_{H_2O}=0.765\cdot M_{H_2O}=13.8\;g$
$m_{Q}=m_{C_2H_6O}+m_{H_2O}=24.6\;g$
$\%_{C_2H_6O}=\frac{m_{C_2H_6O}\cdot 100}{m_{Q}}=43.9\%$
