Sửa: tính $V_{dd\,rượu}$
$n_{C_2H_4}=\frac{4,48}{22,4}=0,2(mol)$
$C_2H_4+H_2O\xrightarrow{axit}C_2H_5OH$
Theo PT: $n_{C_2H_5OH}=n_{C_2H_4}=0,2(mol)$
$\to m_{C_2H_5OH}=0,2.46=9,2(g)$
$\to V_{C_2H_5OH}=\frac{9,2}{0,8}=11,5(ml)$
$\to V_{dd\,rượu}=\frac{11,5.100}{23}=50(ml)$