\(n_{KCl}=n_{MgCl_2}=a\left(mol\right)\\ m_{muối.khan}=m_{KCl}+m_{MgCl_2}=74,5a+95a=169,5a\\ \Leftrightarrow6,78=169,5a\\ \Leftrightarrow a=0,04\\ \Rightarrow n_{H_2O}=\dfrac{11,1-6,78}{18}=0,24\left(mol\right)\\ x=\dfrac{n_{H_2O}}{a}=\dfrac{0,24}{0,04}=6\)