# How much water must be added to 1500 mL of a 6.0 mol/L CaCl_2 solution to make the concentration of the resulting solution 2.0 mol/L?

Jan 26, 2017

You need to add another $\text{3 litres}$ of water........

#### Explanation:

The defining relationship for $\text{concentration}$ is,

$\text{Concentration}$ $=$ $\text{Moles of stuff"/"Volume of solution}$, now typically the solvent is water, but it could be other materials, even a gas.

Here we have a molar quantity of $6.0 \cdot m o l \cdot {L}^{-} 1$, and a $1500 \cdot m L$ volume. We wish to dilute this concentration by a THIRD to $2.0 \cdot m o l \cdot {L}^{-} 1$.

Now clearly, we cannot get the calcium salt out of solution, but we can add more solvent, more water to dilute it.

And here we can use the old relationship,

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$ where $C = \text{concentration}$, and $V = \text{volume}$.

So we solve for ${V}_{2} = \frac{{C}_{1} {V}_{1}}{C} _ 2 = \frac{1500 \cdot m L \times 6.0 \cdot m o l \cdot {L}^{-} 1}{2.0 \cdot m o l \cdot {L}^{-} 1} = 4500 \cdot m L$.

And since we began with $1500 \cdot m L$, another $3000 \cdot m L$ are required.