What is the molarity of chloride in a solution containing 0.084 M CaCl2?

1 Answer

  • 9 years ago
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    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

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