$*)50 ml dd X:\begin{cases}V_1:C_2H_5OH \\ V_2:H_2O\\V_1+V_2=50 ml \end{cases}\Rightarrow \begin{cases}n_{C_2H_5OH}=\frac{0,8V_1}{46} mol \\ n_{H_2O}=\frac{V_2}{18} \end{cases}$$*)n_{H_2}=\frac{15,68}{22,4}=0,7 mol$
$\Rightarrow \frac{n_{C_2H_5OH}+n_{H_2O}}{2}=0,7 mol\Rightarrow \begin{cases}\frac{0,8V_1}{2.46}+\frac{V_2}{2.18}=0,7 mol \\ V_1+V_2=50 ml \end{cases}\Rightarrow V_1\simeq 36 ml$
$\Rightarrow \frac{36.100}{50}=72^o$