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Topic: Problem of the week - 30/07/2012  (Read 16591 times)

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Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #15 on: August 14, 2012, 04:17:14 AM »
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol

No. But I am not able to reproduce the correct answer either  :'(, and as I calculated it in the Dalmatino bar using a napkin, I have no notes to consult.

Trick is, 10.2 g is not a weight of the added gas - it is the increase of the weight, so you should take buoyancy into account. As the volume is 1.22 L, and the air density is around 1.2 g/L, buoyancy is around 1.22*1.2=1.46 g - so the weight of the gas is around 10.2+1.4=11.6 g. That gives molar mass of around 230.

It should be 15.05-1.22*1.2=13.6 g, I am changing the wording of the original question. Sorry about that.
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #16 on: August 14, 2012, 12:31:26 PM »
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol

No. But I am not able to reproduce the correct answer either  :'(, and as I calculated it in the Dalmatino bar using a napkin, I have no notes to consult.

Trick is, 10.2 g is not a weight of the added gas - it is the increase of the weight, so you should take buoyancy into account. As the volume is 1.22 L, and the air density is around 1.2 g/L, buoyancy is around 1.22*1.2=1.46 g - so the weight of the gas is around 10.2+1.4=11.6 g. That gives molar mass of around 230.

It should be 15.05-1.22*1.2=13.6 g, I am changing the wording of the original question. Sorry about that.


no worries  :)


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