DAT Achiever GC Q41

[babowc]

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Any idea where the 0.15mol Cl- in 10cm^3 solution came from?
I get everything else but that...

Thanks guys!

 

[Ferneezy]

no idea where they came up with that. i would skip ahead a little with the understanding that you know chloride would react with silver plus one to one, so i would only calc the number of moles of silver and say that is the number of moles of chloride in my 10ml titration. The rest of the problem works itself out.

 

[HandsOnPractice]

What I did was found the moles of AgNO3 which should be 3X10^-3 moles. Then I went and found the molarity of the unknown chloride solution and found that it was 2X10^-2/2X10^-1 which gave me a 0.1M for the solution. Then I figured out the moles of the unknown chloride solution that would react because they told us only 10ml of the solution would participate in the reaction. When I did that I got 1X10^-1 * 1X10^-2=1X10^-3 then I looked at my mole to mole ratios and because the AgNO3 had three times as many moles I knew I would need that same amount of moles from my unknown chloride to react completely. Looking at the ratio it was 3:1 thus I need a chloride that would give me 3 parts chloride per each part reactant. Thus the only answer possible was AlCl3.

Not sure if that made sense. Hopefully it helped!

 

[hl4jn]

I don't know where you're not understanding so I'll break it down really simply...

You dissolved a compound with the formula XClx with X being the other atom and x being the charge of that atom (like AlCl3 = Al3+ and Cl1-...X is Al and x is 3+)

You know that Ag+ and Cl- react in a 1:1 ratio to form AgCl (s):
Ag+(aq) + Cl-(aq) --> AgCl (s)

So from here, it's all stoichiometry:

20ml of Ag+ x .15 mol/1000 ml of Ag+ = .003 mols of Ag+
.003 mols of Ag+ == .003 mols of Cl- since they're in a 1:1 ration

That means there was .003 mols of Cl- in that 10ml solution of the XCl
.003 mols of Cl-/10ml solution = __ moles of Cl-/200ml solution?
Simple proportion: 200*.003/10

So now you know that there is .06 moles of Cl- in the entire solution.

Let's backtrack a little...
If you have 1 mole of H2O compound, you have 2 moles of H and 1 mole of O
If you have 2 moles of CH4, you have 2 moles of C and 8 moles of H
Understand?

In other words A moles of XyYx = A*y moles of X and A*x moles of Y
*Sorry if that's confusing. I don't know how else to put it.

So if back to the problem.
You have .02 moles of the compound (A) and you have .06 moles of Cl-
.02 * what # gives you .06? 3! So the compound has to be [blahblah]Cl3
This is what they were getting to with the ratio part.

 

[babowc]

I don't know where you're not understanding so I'll break it down really simply...

You dissolved a compound with the formula XClx with X being the other atom and x being the charge of that atom (like AlCl3 = Al3+ and Cl1-...X is Al and x is 3+)

You know that Ag+ and Cl- react in a 1:1 ratio to form AgCl (s):
Ag+(aq) + Cl-(aq) --> AgCl (s)

So from here, it's all stoichiometry:

20ml of Ag+ x .15 mol/1000 ml of Ag+ = .003 mols of Ag+
.003 mols of Ag+ == .003 mols of Cl- since they're in a 1:1 ration

That means there was .003 mols of Cl- in that 10ml solution of the XCl
.003 mols of Cl-/10ml solution = __ moles of Cl-/200ml solution?
Simple proportion: 200*.003/10

So now you know that there is .06 moles of Cl- in the entire solution.

Let's backtrack a little...
If you have 1 mole of H2O compound, you have 2 moles of H and 1 mole of O
If you have 2 moles of CH4, you have 2 moles of C and 8 moles of H
Understand?

In other words A moles of XyYx = A*y moles of X and A*x moles of Y
*Sorry if that's confusing. I don't know how else to put it.

So if back to the problem.
You have .02 moles of the compound (A) and you have .06 moles of Cl-
.02 * what # gives you .06? 3! So the compound has to be [blahblah]Cl3
This is what they were getting to with the ratio part.

Ah, okay..
I got everything but I just didn't understand where the "0.15mol Cl- in 10cm^3 solution" came from.

Now I understand it, I knew it was a simple stoichiometry problem, but got hung up on that other part.

Thanks guys!👍👍👍