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Topic: Amount in mol.  (Read 2403 times)

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Offline RobertT

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Amount in mol.
« on: January 21, 2010, 12:33:02 AM »
I recently stumbled upon a question that I didn't really understand how to answer.

For a 0.20 M solution of potassium sulfate, K2SO4, calculate the amount, in mol, of:
a) Potassium ions, K+
b) Sulfate ions, SO42-
c)Oxygen atoms

I first thought that you would need to do n=C x V but with no Volume given i was unsure of how to answer it
I did decide to convert 0.20M/0.20mol L-1 into g L-1 = 0.20M x 174gMol-1 which gave a result of 34.86 g L-1. This is where i am stuck, I don't know what do do from here =/ Any help is good.
Thanks again, appreciate all the effort.

Offline Ligander

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Re: Amount in mol.
« Reply #1 on: January 21, 2010, 01:40:18 AM »
Take one litre of your solution,  then calculate how many potassium ions etc. you have in your beaker.

Offline Borek

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Re: Amount in mol.
« Reply #2 on: January 21, 2010, 03:01:38 AM »
Without given volume there is no answer to the question. Do like Ligander suggests.
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Offline RobertT

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Re: Amount in mol.
« Reply #3 on: January 21, 2010, 03:50:51 AM »
Take one litre of your solution,  then calculate how many potassium ions etc. you have in your beaker.
Without given volume there is no answer to the question. Do like Ligander suggests.

Ahh, okay, Thanks for the help guys. The solution's for a, b and c are 0.10mol, 0.0500mol and 0.0500mol respectively.

If you consider the volume to be 1 litre, then
a) 0.4 mol K+ ions
b) 0.20 mol SO4 2- ions
c) 0.8 mol O atoms

I think the Book made an error, currently using Heinemann: Chemistry 2. Thanks again for the help, I'll put it down as 1L and continue with the questions. Thanks again for the help.


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