Question

Calculate the mass of `"CuSO"_4 * 5"H"_2"O"` needed to prepare `"100.00 mL"` of a `"0.05000-mol/L"` solution?

Answer 1

Here's what I got.

Explanation:

For starters, you know that `"0.05000-mol L"^(-1)` solution of copper(II) sulfate contains `0.05000` moles of copper(II) sulfate, the solute, for every `"1 L" = 10^3 quad "mL"` of the solution.

This means that your sample must contain

`100.00 color(red)(cancel(color(black)("mL solution"))) * "0.05000 moles CuSO"_4/(10^3 color(red)(cancel(color(black)("mL solution")))) = "0.005000 moles CuSO"_4`

Now, notice that every mole of copper(II) sulfate pentahydrate contains

• `1` mole of anhydrous copper(II) sulfate, `"CuSO"_4`
• `5` moles of water of hydration, `"H"_2"O"`

This means that in order for you to deliver `0.005000` moles of anhydrous copper(II) sulfate to the solution, you must deliver `0.005000` moles of copper(II) sulfate pentahydrate.

In addition to the `0.005000` moles of anhydrous copper(II) sulfate, this sample will contain

`0.005000 color(red)(cancel(color(black)("moles CuSO"_4 * 5"H"_2"O"))) * ("5 moles H"_ 2"O")/(1color(red)(cancel(color(black)("mole CuSO"_4 * 5"H"_2"O")))) = "0.02500 moles H"_2"O"`

So the total mass of the copper(II) sulfate pentahydrate will be equal to the mass of `0.005000` moles of anhydrous copper(II) sulfate and the mass of `0.02500` moles of water of hydration.

`0.005000 color(red)(cancel(color(black)("moles CuSO"_4))) * "159.609 g"/(1color(red)(cancel(color(black)("mole CuSO"_4)))) = "0.798045 g"`

`0.02500 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.450375 g"`

You can thus say that the mass of copper(II) sulfate pentahydrate that will deliver `0.005000` moles of anhydrous copper(II) sulfate to your solution will be

`"0.798045 g + 0.450375 g" = color(darkgreen)(ul(color(black)("1.248 g")))`

The answer is rounded to four sig figs, the number of sig figs you have for the molarity of the solution.