Question

Question #8efd1

Answer 1

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`