Question

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`?

Answer 1

`"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 `"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 `"4.49 g"` of copper(II) chloride, the solute.

`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 `"4.23 mL"` of `"7.90 M"` copper(II) chloride solution to a total volume of `"51.5 mL"`, you will have a solution that contains `"4.49 g"` of copper(II) chloride.