\(n_{HCl}=\dfrac{V}{22,4}=\dfrac{8.96}{22,4}=0,4\left(mol\right)\)
\(m_{HCl}=n.M=0,4.36,5=14,6\left(g\right)\)
\(m_{dd}=m_{ct}+m_{dm}=14,6+185,5=200,1\left(g\right)\)
\(C_{\%\left(HCl\right)}=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{14,6}{200,1}.100=7,3\left(\%\right)\)