Question
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?
Thanks!
Thanks!
Answers
(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.
0.075 mols/0.425 L = ?
Since there are two Cl^- in a mol of CaCl2, the (Cl^-) = twice that.
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