Given mole fraction of ethanol \(=0.040\)
\(\begin{aligned}
&\text { i.e., } \mathrm{X}_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=\frac{\mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}}{\mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}+\mathrm{n}_{\left(\mathrm{H}_2 \mathrm{O}\right)}} \\
&=0.040
\end{aligned}\)
Now we have to find number of moles of ethanol in \(1 \mathrm{~L}\) of the solution which is nearly \(=1 \mathrm{~L}\) of water (because solution is dilute)
No. of moles in \(1 \mathrm{~L}\) water \(=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) \(=55.55\) moles
Substituting \(\mathrm{n}\left(\mathrm{H}_2 \mathrm{O}\right)=55.55\) in eqn (i), we get
\(\begin{aligned}
&\frac{\mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}}{\mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}+55.55}=0.040 \\
&0.040 \mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}+55.55 \times 0.040=\mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)} \\
&\text { or } 0.96 \mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}=55.55 \times 0.040 \\
&\text { or } \mathrm{n}_{\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)}=2.31 \mathrm{~mol}
\end{aligned}\)
Hence, molarity of the solution \(=2.31 \mathrm{M}\)