A solution of  36%  by mass water and  64%  by mass acetaldehyde $\left(C{H}_{3}CHO\right)$ . Mole fraction of acetaldehyde is

36%  $H_{2}O$
36g  of $H_{2}O$   in 100g  of solution
64%  $C{H}_{3}CHO$
64g  of  $C{H}_{3}CHO$  in 100g  of solution
 $\begin{array}{l}{n}_{C{H}_{3}CHO}=\frac{64}{44}{n}_{H_{2}O}=\frac{36}{18}\\ n_{C{H}_{3}CHO}=1.45n_{H_{2}O}=2\\ X_{C{H}_{3}CHO}=\frac{1.45}{1.45+2}\\ X_{C{H}_{3}CHO}=0.42\end{array}$