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Topic: Problem of the week - 27/05/2013  (Read 21501 times)

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Offline Borek

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Problem of the week - 27/05/2013
« on: May 27, 2013, 06:39:20 AM »
What is the density of the sulfuric acid solution (g/mL) if its percent w/w concentration is 60.16% and w/v percent concentration is 90.24%?
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Offline Schrödinger

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Re: Problem of the week - 27/05/2013
« Reply #1 on: May 27, 2013, 11:56:20 PM »
bar cbvag svir g/mL?
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Offline curiouscat

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Re: Problem of the week - 27/05/2013
« Reply #2 on: May 28, 2013, 12:45:47 AM »
[tex]
\LARGE{\sqrt[3]{3.375}}
[/tex]

g/ml

Offline Borek

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Re: Problem of the week - 27/05/2013
« Reply #3 on: May 28, 2013, 11:57:43 AM »
Have you used density tables?
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Offline Rutherford

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Re: Problem of the week - 27/05/2013
« Reply #4 on: May 28, 2013, 12:25:42 PM »
I didn't.
ω(mass share)=mH2SO4/msolution=0.6016
φ(volume share)=VH2SO4/Vsolution=0.9024
Dividing:
ω/φ=Vsolution/msolution=0.6667
Vsolution=0.6667msolution
ρsolution=msolution/Vsolution=msolution/0.6667msolution=1.5gcm-3

Offline curiouscat

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Re: Problem of the week - 27/05/2013
« Reply #5 on: May 28, 2013, 12:48:19 PM »
Have you used density tables?

Nope.

Offline Schrödinger

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Re: Problem of the week - 27/05/2013
« Reply #6 on: May 28, 2013, 12:48:50 PM »
Have you used density tables?
:O No!

It's a fairly simple question. If 100 g of the soln contains 60.16 g of H2SO4 and 100 mL of the solution contains 90.24 g of H2SO4, all one needs to find out is how many mL of soln corresponds to 60.16 g H2SO4, because this volume of soln which corresponds to 60.16 g of H2SO4, also weighs 100 g (given).
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Offline Borek

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Re: Problem of the week - 27/05/2013
« Reply #7 on: May 28, 2013, 01:33:40 PM »
I know you don't have to. I was wondering how to word the question - whether to name the compound or not. If it is named, it suggests it is a specific case, not a general thing.
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Offline Big-Daddy

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Re: Problem of the week - 27/05/2013
« Reply #8 on: May 30, 2013, 03:51:38 PM »
What's the justification for V[H2SO4]/m[H2SO4]=1?

Only then ρ[Solution]=Volume Share/Mass Share=1.500 g/cm3. Otherwise Volume Share/Mass Share=ρ[Solution]*V[H2SO4]/m[H2SO4].

Offline curiouscat

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Re: Problem of the week - 27/05/2013
« Reply #9 on: May 30, 2013, 04:06:43 PM »
What's the justification for V[H2SO4]/m[H2SO4]=1?

Only then ρ[Solution]=Volume Share/Mass Share=1.500 g/cm3. Otherwise Volume Share/Mass Share=ρ[Solution]*V[H2SO4]/m[H2SO4].

density of solution is always equal to mass / volume.

The rest of what you wrote didn't make any sense to me. Especially not what "share" applies to.

Your equations don't even make dimensional sense.

Offline Big-Daddy

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Re: Problem of the week - 27/05/2013
« Reply #10 on: May 30, 2013, 04:24:55 PM »
The rest of what you wrote didn't make any sense to me. Especially not what "share" applies to.

I was looking at Raderford's derivation as it is the only one on the page.

ω(mass share)=mH2SO4/msolution=0.6016
φ(volume share)=VH2SO4/Vsolution=0.9024

So there are the definitions of Mass Share and Volume Share.

Dividing:
ω/φ=Vsolution/msolution=0.6667

Look at the expressions for ω and φ; clearly ω/φ=mH2SO4·Vsolution/(msolution·VH2SO4). Only if mH2SO4/VH2SO4=1 does this reduce to ω/φ=Vsolution/msolution. What's the justification for that?

Offline curiouscat

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Re: Problem of the week - 27/05/2013
« Reply #11 on: May 30, 2013, 04:27:36 PM »
He has a Typo. V should be m.

Offline Big-Daddy

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Re: Problem of the week - 27/05/2013
« Reply #12 on: May 30, 2013, 05:22:17 PM »
He has a Typo. V should be m.

Ah I see. w/v. It's not volume fraction we want but mass concentration, so p[H2SO4]=m[H2SO4]/V[solution]. Got it now, thanks for the help.

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