Explanation

(a)
Number of moles of ethyl alcohol = $\frac{138}{46}$= 3
Number of moles of water =$\frac{72}{18}$ =4
$X_{C_{2}H_{5}OH} = \frac{3}{3 + 4} = \frac{3}{7}\ \ \ \ \$
$\ \ \ \ \ \ \ \ X_{H_{2}O} = \frac{4}{3 + 4} = \frac{4}{7}$
$\frac{\ X_{C_{2}H_{5}OH}}{X_{H_{2}O}} = \frac{\frac{3}{7}}{\frac{4}{7}}$ = $\frac{3}{4}$