What is the concentration of chloride ions (in) in a solution prepared by adding 0.075 moles of calcium chloride to 425 mL water,assuming no volume change up on mixing?

Question

Thanks!

Answer 1

(CaCl2) = M = moles/L soln.
0.075 mols/0.425 L = ?
Since there are two Cl^- in a mol of CaCl2, the (Cl^-) = twice that.