Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is $0.040$ (assume the density of water to be $1 \, g/mL$).

(N/A) Mole fraction of $C_2H_5OH = \frac{\text{Number of moles of } C_2H_5OH}{\text{Number of moles of solution}}$
$0.040 = \frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + n_{H_2O}}$ ...........$(I)$
Number of moles present in $1 \, L$ water:
$n_{H_2O} = \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.55 \, mol$
Substituting the value of $n_{H_2O}$ in equation $(I)$:
$\frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + 55.55} = 0.040$
$n_{C_2H_5OH} = 0.040 \, n_{C_2H_5OH} + 2.222$
$0.96 \, n_{C_2H_5OH} = 2.222 \, mol$
$n_{C_2H_5OH} = \frac{2.222}{0.96} \, mol = 2.314 \, mol$
Since the volume of the solution is approximately $1 \, L$ (assuming dilute solution density $\approx 1 \, g/mL$):
Molarity $= \frac{2.314 \, mol}{1 \, L} = 2.314 \, M$