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

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.