# What volume of 133 mL, 7.90 M #CuCl_2# solution should you dilute in order to have 51.5 mL of diluted solution that contains 4.49 g #CuCl_2#?

##### 1 Answer

#### Explanation:

The first thing to do here is to use the *molarity* and the *volume* of the initial solution to figure out how many **grams** of copper(II) chloride it contains.

#133 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("7.90 moles CuCl"_2)/(1color(red)(cancel(color(black)("L solution")))) = "1.051 moles CuCl"_2#

To convert this to *grams*, use the compound's **molar mass**

#1.051 color(red)(cancel(color(black)("moles CuCl"_2))) * "134.45 g"/(1color(red)(cancel(color(black)("mole CuCl"_2)))) = "141.31 g CuCl"_2#

Now, you know that the diluted solution must contain **dilute** a solution, you increase the amount of solvent while keeping the amount of solute **constant**.

This means that you must figure out what volume of the **initial solution** will contain

#4.49 color(red)(cancel(color(black)("g"))) * "133 mL solution"/(141.32color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("4.23 mL solution")))#

The answer is rounded to three **sig figs**.

You can thus say that when you dilute