Question #8efd1

1 Answer
Jul 27, 2017

Answer:

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#