\(Cl_2+H_2\rightarrow2HCl\)
\(n_{Cl2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,3\left(mol\right)\)
\(m_{dd\left(B\right)}=0,3.36,5+100=110,95\left(g\right)\)
\(\Rightarrow V_{dd\left(B\right)}=\frac{110,95}{1,132}=98\left(ml\right)\)
\(C\%_{HCl}=\frac{0,3.36,5}{110,95}.100\%=9,87\%\)
\(\Rightarrow CM_{HCl}=\frac{0,3}{0,098}=3,06M\)