Question #8efd1

1 Answer
Al E.
Jul 27, 2017

The molarity of chloride ions suspended in the aqueous solution is 2.00M.

Explanation:

Here, we're looking to find the molarity of Cl^- ions in the solution.

27.75g * (CaCl_2)/(110.98g) * 1/(.250L) = 1.00M

The foregoing conversion helps us realize the molarity of the entire solution is 1.00M, but remember, there are 2 moles of Cl^- per one mole of the soluble compound, so:

(1.00mol)/L * (2Cl^-)/(CaCl_2) = 2.00M

Related questions
See all questions in Molarity
Impact of this question
1384 views around the world
You can reuse this answer
Creative Commons License