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

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SUMMARY

The calculation of moles and ions in an aqueous lithium nitrate (LiNO3) solution involves determining the number of moles based on the concentration and volume of the solution. For a solution with 6.96 x 1021 formula units per liter and a volume of 3.26 L, the correct calculation yields approximately 1.16 x 10-2 moles of Li and NO3 ions, each present in a 1:1 ratio. The user initially miscalculated by not incorporating the volume correctly, leading to confusion regarding the total ion count. The final concentration of the solution is approximately 0.003 mol/dm3.

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brandon1
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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
 
Last edited:
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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?
 
Last edited:

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