\(n_{HCl}=\frac{6,2}{22,4}=\frac{31}{112}\left(mol\right)\)
\(\Rightarrow C_M=\frac{\frac{31}{112}}{\frac{89,05}{1000}}=3,1\left(M\right)\)
\(m_{HCl}=\frac{31}{112}.36,5=9,753\left(g\right)\)
\(\Rightarrow C\%=\frac{9,753}{9,753+89,05.1}.100\%=9,87\%\)
\(\Rightarrow D\left(dd\right)=\frac{9,753+89,05}{89,05}=1,11\)