# Chemistry Question on moles of ions

How many moles of ions are present in aqueous solutions prepared by dissolving 10.00 g of the following compounds in water to make 4.35 L of solution?

(a) cobalt(III) chloride
(b) aluminum carbonate
(c) potassium permanganate
(d) strontium hydroxide

I first converted the 10.00 g into moles, and divided by the molar mass of each of the compounds, and multiplied the result by 4.35 L. But I am not obtaining the correct answer.

Any help is appreciated

Thanks

Staff Emeritus
Gold Member
plugpoint said:
I first converted the 10.00 g into moles, and divided by the molar mass of each of the compounds, and multiplied the result by 4.35 L.

Can you show how you did that for the first compound? It will make it easier to see where you're going wrong.

Thanks.

Staff Emeritus
Gold Member
Upon closer inspection, I have a question.

plugpoint said:
and multiplied the result by 4.35 L.

Why would you do that? If you convert from grams to moles, multiplying by liters is only going to screw up the units.

Ok so we have cobalt(III) chloride or $CoCl_{3}$. Its molar mass is $58.93 + 3(35.34) = 164.95$. 10 g of this stuff is thus 0.06 moles. All I know know is that $M = \frac{mol sol}{L solv}$

Gold Member
Yah, $M = \frac{moles}{volume}$

Staff Emeritus
Gold Member
plugpoint said:
Ok so we have cobalt(III) chloride or $CoCl_{3}$. Its molar mass is $58.93 + 3(35.34) = 164.95$. 10 g of this stuff is thus 0.06 moles.

OK, that's a big help. I can clearly see exactly why you are not answering the question.

Look at the dissociation reaction:

[tex]CoCl_3(s)\longrightarrow Co^{3+}(aq)+3Cl^{1-}(aq)[/itex]

Now the question asks you how many moles of ions are in each solution. So once you have the number of moles of $CoCl_3$, you need to use stoichiometry to get the number of moles of each ionic species. Then, since the question asks for the number of moles of ions in the solution, you'll have to add up the results.

Same goes for the other compounds.

All I know know is that $M = \frac{mol sol}{L solv}$

You don't need this at all.

Staff Emeritus
Ok so we have cobalt(III) chloride or $CoCl_{3}$. Its molar mass is $58.93 + 3(35.34) = 164.95$. 10 g of this stuff is thus 0.06 moles. All I know know is that $M = \frac{mol sol}{L solv}$