$n_{HCl}=\dfrac{22,4}{22,4}=1(mol)$
$\to m_{HCl}=1.36,5=36,5g$
$3,35g$ nước cất hoà tan tối đa $36,5g$ $HCl$
$\to 100g$ nước hoà tan tối đa $\dfrac{36,5.100}{3,35}=1089,5g$
Vậy $S_{HCl}=1089,5g$
$C\%_{\text{bão hoà}}=\dfrac{36,5.100}{36,5+3,35}=91,6\%$