\(n_{CaCl_2.6H_2O}=\dfrac{25}{219}\left(mol\right)\)
=> \(n_{CaCl_2}=\dfrac{25}{219}\left(mol\right)\)
\(C_M=\dfrac{\dfrac{25}{219}}{0,312}=\dfrac{3125}{8541}M\)
\(m_{dd}=1,09.312=340,08\left(g\right)\)
\(C\%=\dfrac{\dfrac{25}{219}.111}{340,08}.100\%=3,726\%\)