# A solution was prepared by dissolving 66.6 g of calcium chloride in enough water to make 175 mL of solution. This solution was then further diluted to a total of 550 mL with water. What was the concentration of the final solution?

Apr 16, 2016

$\text{Concentration "="Moles of solute"/"Volume of solution}$, $\text{approx. } 1 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$C$ $=$ $\frac{n}{V}$, where $n = \frac{66.6 \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.600$ $m o l$.

The final volume of the solution is $550 \times {10}^{-} 3 \cdot L$

So $\text{Concentration with respect to } C a C {l}_{2} = \frac{0.600 \cdot m o l}{550 \times {10}^{-} 3 \cdot L}$ $=$ ??

What is the concentration with respect to $C {a}^{2 +} \left(a q\right)$, to $C {l}^{-} \left(a q\right)$?