Gọi $x$, $y$ là số mol $C_2H_5OH$, $H_2O$
$\to 46x+18y=10,04$ $(1)$
$n_{H_2}=\dfrac{3,808}{22,4}=0,17(mol)$
$\to n_{OH}=2n_{H_2}=0,17.2=0,34(mol)$
$\to x+y=0,34$ $(2)$
$(1)(2)\to x=0,14; y=0,2$
$m_{C_2H_5OH}=46x=6,44g\to V_{C_2H_5OH}=\dfrac{6,44}{0,8}=8,05ml$
$m_{H_2O}=18y=3,6g\to V_{H_2O}=3,5ml$
Độ rượu:
$D=\dfrac{8,05.100}{8,05+3,5}=69^o$