# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on May 27, 2013, 06:39:20 AM

Title: Problem of the week - 27/05/2013
Post by: Borek 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%?
Title: Re: Problem of the week - 27/05/2013
Post by: Schrödinger on May 27, 2013, 11:56:20 PM
bar cbvag svir g/mL?
Title: Re: Problem of the week - 27/05/2013
Post by: curiouscat on May 28, 2013, 12:45:47 AM
$$\LARGE{\sqrt[3]{3.375}}$$

g/ml
Title: Re: Problem of the week - 27/05/2013
Post by: Borek on May 28, 2013, 11:57:43 AM
Have you used density tables?
Title: Re: Problem of the week - 27/05/2013
Post by: Rutherford 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
Title: Re: Problem of the week - 27/05/2013
Post by: curiouscat on May 28, 2013, 12:48:19 PM
Have you used density tables?

Nope.
Title: Re: Problem of the week - 27/05/2013
Post by: Schrödinger 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).
Title: Re: Problem of the week - 27/05/2013
Post by: Borek 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.
Title: Re: Problem of the week - 27/05/2013
Post by: Big-Daddy 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].
Title: Re: Problem of the week - 27/05/2013
Post by: curiouscat 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.
Title: Re: Problem of the week - 27/05/2013
Post by: Big-Daddy 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?
Title: Re: Problem of the week - 27/05/2013
Post by: curiouscat on May 30, 2013, 04:27:36 PM
He has a Typo. V should be m.
Title: Re: Problem of the week - 27/05/2013
Post by: Big-Daddy 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.