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

1. 1m 2. 2M 3. 0.5M 4. 0.2M

Concentration in molarity , M = moles of CaCl₂ /volume in L
And moles of CaCl₂ = mass/molar mass  mass = 11.1g  molar mass =  111g/mol  (40 + 2 X 35.5)
Moles of CaCl₂ =  11.1/111= 0.1 mol
c = 0.1/0.1 = 1M  (Volume of solution = 100ml = 0.1L)

Now, CaCl₂ dissociates in solution to form one mole of calcium cations and 2 moles of chloride anions
CaCl₂ $\iff$ Ca²⁺  +  2Cl^-

So, concentration of chloride ion is twice that of CaCl₂ and is 2M  (option 2)