# To what volume should you dilute 150 mL of an 8.00 M CuCl2 solution so that 55.0 mL of the...

## Question:

To what volume should you dilute 150 mL of an 8.00 M {eq}CuCl_2 {/eq} solution so that 55.0 mL of the diluted solution contains 5.9 g of {eq}CuCl_2 {/eq}?

## Dilution of Solutions:

During dilution, what we are doing is we're adding enough pure solvent, which most of the time is just water, to a given volume of concentrated solution to achieve our desired concentration of the solution. Although the volume changed, the amount of solute present before and after dilution is the same.

We are given:

• {eq}M_1 {/eq} = 8.00 M CuCl2
• {eq}V_1 {/eq} = 150 mL

From the given information, we need to determine to what volume we need to dilute the 150 mL of 8.00 M CuCl2 solution so that 55.0 mL of the diluted solution will contain 5.9 g of CuCl2.

Solution:

The aliquot of the diluted solution with a volume of 55.0 mL contains 5.9 g of CuCl2. Using this information, we need to calculate the molarity of the diluted solution. The molar mass of CuCl2 is 134.45 g/mol. Solving for the molarity of the diluted solution, we get,

{eq}\begin{align} M_2 &= \rm \left(\dfrac{5.9\ g\ CuCl_2}{55.0\ mL} \right) \left(\dfrac{1\ \text{mol}\ CuCl_2}{134.45\ g\ CuCl_2} \right) \left(\dfrac{1000\ mL}{1\ L} \right)\\[2ex] &= 0.80\ \text{mol/L}\\[2ex] &= 0.80\ \text{M} \end{align} {/eq}

Therefore, the molarity of the diluted solution is 0.80 M. To determine the volume of the diluted solution, we will use the dilution formula.

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 {/eq} and {eq}V_1 {/eq} are the molarity and volume of the concentrated solution, while,
• {eq}M_2 {/eq} and {eq}V_2 {/eq} are the molarity and volume of the diluted solution.

Rearranging the equation to get {eq}V_2 {/eq} gives us:

{eq}V_2 = \dfrac{M_1V_1}{M_2} {/eq}

Substituting the known values in the equation, we get:

{eq}\begin{align} V_2 &= \dfrac{M_1V_1}{M_2}\\[2ex] &= \rm \dfrac{(8.00\ M)(150\ mL) }{0.80\ \text{M}} \end{align} {/eq}

{eq}\boxed{\mathbf{V_2 = 1500\ \text{mL}}} {/eq}

Therefore, we need to dilute the 150 mL of 8.00 M solution to 1500 mL. This means that we need to add 1350 mL of pure water.