11.1 g. of CaCl₂ is present in 100 ml of the aqueous solution. The chloride ion concentration is

Molar mass of CaCl₂​=111g

Number of moles =$\frac{11.1}{111}$​=0.1mole.

Now, $\underset{0.1\mathrm{mole}}{{\mathrm{CaCl}}_2}\left(\mathrm{aq}\right)⟶\underset{0.1\mathrm{mol}}{{\mathrm{Ca}}^{+2}}\left(\mathrm{aq}\right)+\underset{0.2\mathrm{mol}}{2{\mathrm{Cl}}^{-}}$

Molarity$=\frac{\mathrm{number} \mathrm{of} \mathrm{moles}}{\mathrm{volume}}=\frac{0.2}{0.1}=2\mathrm{M}$