$x_{\mathrm{C}_2\mathrm{H}_2\mathrm{OH}}$ = $\frac{n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}}{n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}+n_{\mathrm{H}_2\mathrm{O}}}$ = 0.40 (Given) ... (i) The aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1 L of water (because solution is dilute) Number of moles of water in 1 L of water = $\frac{1000\,\mathrm{g}}{18\,\mathrm{g}\,\mathrm{mol}^{-1}}$ = 55.55 moles Substituting $n_{\mathrm{H}_2\mathrm{O}}$ = 55.55 in eqn (i), we get $\frac{n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}}{n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}+55.55}$ = 0.040 or 0.96 $n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}$ = 55.55 × 0.040 or $n_{\mathrm{C}_2\mathrm{H}_5\mathrm{OH}}$ = 2.31 mol Hence, molarity of the solution = 2.31 M.