How do you calculate moles and ions in an aqueous lithium nitrate solution?

[brandon1]

How many moles and numbers of ions of each type are present in the following aqueous solutions?

3.26 L of a solution containing 6.96 x 10^21 formula units of lithium nitrate per liter

I got 1.16e-2 mols of Li and NO3 as well as 6.99e21 ions for each. The software I am using (webassign) says it is incorrect. What am I doing wrong?

First, I found the number of moles of the total was 1.16e-2 moles. Since the two ions are in a 1:1 ratio, should they not be the same as the total? Or am I supposed to use the 3.26L in there somewhere?EDIT: I found what I did wrong. I didn't see the per liter at the end

 

[3trQN]

3.26 L = 3.26 $$dm^3$$

There is approx $$6.02 \times10^{23}$$ "formula units" in a mole

$$\frac{6.96 \times10^{21}}{6.02 \times10^{23}}$$

$$\frac{6.96}{6.02} \times 10^{21-23}$$

$$1.16 \times 10^{-2}$$ moles

$$[LiNO_{3}] = \frac{1.16\times10^{-2}}{3.26} mol dm^{-3}$$

Is the answer ~ 0.003?

Do you have to take dissociation constant into account?