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Topic: Degree of dissociation  (Read 5119 times)

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Offline Shadow

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Degree of dissociation
« on: April 27, 2012, 02:39:24 PM »
In a solution of zinc-chloride there are 1.806*1022 chloride ions and 11.56g of salt that didn't dissociate. Calculate the degree of dissociation in that solution. Is it 13%?
« Last Edit: April 27, 2012, 02:51:32 PM by Shadow »

Offline Borek

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Re: Degree of dissociation
« Reply #1 on: April 27, 2012, 04:14:41 PM »
You are on the right track, but you have ignored stoichiometry of the dissociation.
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Offline Shadow

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Re: Degree of dissociation
« Reply #2 on: April 28, 2012, 06:03:52 AM »
Calculated the n of the salt that didn't dissociate and added to it n(Cl-)/2.
Then I divided n(Cl-)/2 with the number I got previously and I got 0.013=13%. What's wrong here?

Offline Shadow

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Re: Degree of dissociation
« Reply #3 on: April 29, 2012, 12:35:23 PM »
How to get 15% here (the right answer)?
ZnCl2 :rarrow: Zn2++2Cl-
I calculated the degree: n(ZnCl2 that dissociated)/n(ZnCl2 at the beginning)*100 and got 13%.
n(ZnCl2 that dissociated) is equal to n(Cl-)/2.
n(ZnCl2 at the beginning) is equal to n(ZnCl2 that dissociated) + n of the undissociated salt.
It can't be a mistake in calculation.

Offline Borek

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Re: Degree of dissociation
« Reply #4 on: April 29, 2012, 04:36:23 PM »
Please post exact numbers you are working with. I am getting a different answer. Doesn't mean I am OK, but we need to compare details to be sure who is right.
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Offline Shadow

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Re: Degree of dissociation
« Reply #5 on: April 30, 2012, 02:45:27 AM »
n(ZnCl2 that dissociated) is equal to n(Cl-)/2, that is 1.806*1022/6*1023 all divided by 2, so it is 0.01505mol.
n(ZnCl2 at the beginning) is equal to n(ZnCl2 that dissociated) + n of the undissociated salt so it is 0.01505+11.56/(65+71)=0.10005mol
The degree is 0.01505/0.10005*100=15% not sure where I made a mistake last time, but I got the right one now. Thanks for help.

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