What is the molarity of chloride in a solution containing 0.084 M CaCl2?
- Robert DLv 79 years agoFavorite Answer
The CaCl2 is completely soluble in water so for each formula unit of CaCl2 you obtain 1 Ca^2+ ion and 2 Cl^- ions
molarity of Cl^- = (0.084 moleCaCl2/L)(2 mole Cl^-)/(1 mole CaCl2) = 0.168 mole Cl^-/L
The molarity of Cl^- is 0.168 M.
Hope this is helpful to you. JIL HIR