# 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?

Apr 1, 2017

$\text{4.23 mL}$

#### 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 $\text{4.49 g}$ of copper(II) chloride. As you know, when you 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 $\text{4.49 g}$ of copper(II) chloride, the solute.

$4.49 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "133 mL solution"/(141.32color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("4.23 mL solution}}}}$

The answer is rounded to three sig figs.

You can thus say that when you dilute $\text{4.23 mL}$ of $\text{7.90 M}$ copper(II) chloride solution to a total volume of $\text{51.5 mL}$, you will have a solution that contains $\text{4.49 g}$ of copper(II) chloride.