# 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.

## Answer and Explanation: 1

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.

#### Learn more about this topic:

from

Chapter 8 / Lesson 5Want to know how to calculate dilution factor? See dilution equations, the dilution formula, and learn how to dilute acid and how to dilute a solution.