# Question 07829

Nov 3, 2016

$\text{0.60 g}$

#### 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, $\text{Cu"("NO"_3)_2 * 3"H"_2"O}$.

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, $3 \times \text{H"_2"O}$

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")))) #

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{0.60 g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.

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