Question #07829

The trick here is to realize that your solute is anhydrous copper(II) nitrate, #"Cu"("NO"_3)_2#, which you deliver to the solution by dissolving the copper(II) nitrate trihydrate, #"Cu"("NO"_3)_2 * 3"H"_2"O"#.

Calculate the moles of solute present in your solution by using its molarity as a conversion factor

#25 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.10 moles Cu"("NO"_3)_2)/(1color(red)(cancel(color(black)("L solution"))))#

#= "0.0025 moles Cu"("NO"_3)_2#

Now, notice that #1# mole of copper(II) nitrate trihydrate contains

• one mole of anhydrous copper(II) nitrate, #1 xx "Cu"("NO"_3)_2#
• three moles of water of crystallization, #3 xx "H"_2"O"#

This means that in order to deliver #0.0025# moles of anhydrous salt to the solution, you must use #0.0025# moles of hydrate.

To convert this to grams, use the molar mass of the hydrate

#0.0025color(red)(cancel(color(black)("moles Cu"("NO"_3)_2))) * (1color(red)(cancel(color(black)("mole Cu"("NO"_3)_2 * 3"H"_2"O"))) )/(1color(red)(cancel(color(black)("mole Cu"("NO"_3)_2)))) * "241.6 g"/(1color(red)(cancel(color(black)("mole Cu"("NO"_3)_2 * 3"H"_2"O")))) #

#= color(green)(bar(ul(|color(white)(a/a)color(black)("0.60 g")color(white)(a/a)|)))#

The answer is rounded to two sig figs.

So, in order to prepare your solution, take #"0.60 g"# of copper(II) nitrate trihydrate and dissolve the sample in enough water to make the total volume of the solution equal to #"25 mL"#.