How Many Moles of Ions Are in These Aqueous Solutions?

[courtrigrad]

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

 

[quantumdude]

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.

 

[quantumdude]

Upon closer inspection, I have a question.

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.

 

[courtrigrad]

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 $M = \frac{mol sol}{L solv}$

 

[Pengwuino]

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

 

[quantumdude]

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]<br /> <br /> Now the question asks you how many moles of <b>ions</b> 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.<br /> <br /> Same goes for the other compounds.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> All I know know is that $M = \frac{mol sol}{L solv}$ </div> </div> </blockquote><br /> You don't need this at all.[/tex]

 

[Moonbear]

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 $M = \frac{mol sol}{L solv}$

Here's a hint for you: Read again what it is that the question is asking for. Sometimes there is extra information you don't need (or maybe you need it for another part of the question). Also, it is asking for how many moles of ions you have. What extra step might you need to add to account for that?

Edit: Tom was quicker than I was!