Theo đề, ta có: \(n_{HCl}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(\Rightarrow m\)dung dịch HCl \(=m_{HCl}+m_{H_2O}=3,65+46,35=50\left(g\right)\)
\(\Rightarrow C\%\)dung dịch HCl \(=\dfrac{3,65}{50}.100\%=7,3\%\)