Moles of ions in aqueous 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?

 

[Blueshift5]

of the given information. So I had to multiply the moles by 3.26 to get the correct answer.

I would like to clarify that the number of moles and the number of ions in a solution are two different quantities. Moles represent the amount of a substance, while the number of ions represents the number of charged particles present in the solution.

To answer the question, we first need to calculate the number of moles of lithium nitrate in the given solution. This can be done by multiplying the concentration (6.96 x 10^21 formula units per liter) by the volume (3.26 L). This gives us a total of 2.27 x 10^22 formula units of lithium nitrate.

Next, we need to convert this into moles. Since lithium nitrate has a 1:1 ratio of moles to formula units, we can simply divide the number of formula units by Avogadro's number (6.02 x 10^23). This gives us a total of 3.77 x 10^-2 moles of lithium nitrate in the solution.

Now, to calculate the number of ions, we need to consider that lithium nitrate dissociates into two ions in aqueous solution - one lithium ion (Li+) and one nitrate ion (NO3-). This means that the total number of ions in the solution would be twice the number of moles of lithium nitrate. So, the total number of ions in the solution would be 7.54 x 10^-2 ions.

In conclusion, the correct answer would be 3.77 x 10^-2 moles of lithium nitrate and 7.54 x 10^-2 ions (3.77 x 10^-2 Li+ ions and 3.77 x 10^-2 NO3- ions) in the given solution. I hope this clarifies any confusion and helps you understand the concept better.