Calculate the molarity of a solution of ethanol in water

Solution:

Mole fraction of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}=\frac{\text { Number of moles of } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\text { Number of moles of solution }}$

$0.040=\frac{n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}}{n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}+n_{\mathrm{H}_{2} \mathrm{O}}}$ .....(i)

Number of moles present in 1 L water

$n_{\mathrm{H}_{2} \mathrm{O}}=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$

$n_{\mathrm{H}_{2} \mathrm{O}}=55.55 \mathrm{~mol}$

Substituting the value of $n_{\mathrm{H}, \mathrm{O}}$ in equation (1),



$\therefore$ Molarity of solution $=\frac{2.314 \mathrm{~mol}}{1 \mathrm{~L}}$

$=2.314 \mathrm{M}$