# Question #8efd1

Jul 27, 2017

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

#### Explanation:

Here, we're looking to find the molarity of $C {l}^{-}$ ions in the solution.

$27.75 g \cdot \frac{C a C {l}_{2}}{110.98 g} \cdot \frac{1}{.250 L} = 1.00 M$

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

$\frac{1.00 m o l}{L} \cdot \frac{2 C {l}^{-}}{C a C {l}_{2}} = 2.00 M$