Question

Question #07829

Answer 1

`"0.60 g"`

Explanation:

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"`.