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Topic: (%w/w) to (%w/v)  (Read 4642 times)

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Offline defencegrid

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(%w/w) to (%w/v)
« on: July 08, 2017, 02:28:32 AM »
Just have a query about the following question:

How would you prepare 1L of a 1% (w/v) H2SO4 solution from a 98% (w/w) stock of H2SO4?

Density of H2SO4 is 1.84 g/cm³

Molar mass of H2SO4 = 98.07848 g/mol

The answer I get is 5.6ml of 98% H2SO4 and make up to 1L.  I got this answer by converting the concentration of both the stock and final solution to molarity and then doing c1/v1=c2/v2.  Is my answer correct?

My other question is, I don't think I can just do a c1/v1= c2/v2 with the percentages alone, because one is (w/w) and the other is (w/v)? Is this correct?
« Last Edit: July 08, 2017, 06:52:36 AM by Arkcon »

Offline Arkcon

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Re: (%w/w) to (%w/v)
« Reply #1 on: July 08, 2017, 07:17:17 AM »
The answer I get is 5.6ml of 98% H2SO4 and make up to 1L.

I'll check your calc in a minute.

Quote
I got this answer by converting the concentration of both the stock and final solution to molarity and then doing c1/v1=c2/v2.


A good way of doing it.  Query: what is the molarity of 98% w/w sulfuric, and 1% w/v sulfuric?  Different values are published, we won't get the same result, unless we're using the same reference.

Quote
My other question is, I don't think I can just do a c1/v1= c2/v2 with the percentages alone, because one is (w/w) and the other is (w/v)? Is this correct?

I wonder.  For getting the 'bestest' answer for an online homework problem, sure.  For just running a reaction -- you're only 2 % off if you just assume the sulfuric acid is 100% pure.  Will that cause a problem?  I'm not convinced that is the case.  For analytical purposes, you'd be expected to standardize, so what dose it really matter?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline defencegrid

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Re: (%w/w) to (%w/v)
« Reply #2 on: July 08, 2017, 09:52:48 PM »
The molarity I get for of 98% (w/w) sulphuric acid is 18.4M

The molarity I get for 1% (w/v) sulphuric acid is 0.10M

I agree for reactions it doesn't really matter and for analytical purposes you would standardise, but if I'm preparing a solution I like to prepare it as accurately as possible. 

Offline Vidya

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Re: (%w/w) to (%w/v)
« Reply #3 on: July 27, 2017, 10:57:19 AM »
Just have a query about the following question:

How would you prepare 1L of a 1% (w/v) H2SO4 solution from a 98% (w/w) stock of H2SO4?

Density of H2SO4 is 1.84 g/cm³

Molar mass of H2SO4 = 98.07848 g/mol

The answer I get is 5.6ml of 98% H2SO4 and make up to 1L.  I got this answer by converting the concentration of both the stock and final solution to molarity and then doing c1/v1=c2/v2.  Is my answer correct?

My other question is, I don't think I can just do a c1/v1= c2/v2 with the percentages alone, because one is (w/w) and the other is (w/v)? Is this correct?
Your calculations for the molarity of H2SO4 is right.How did you get 0.1 M as the concentration of H2SO4 ???
I think it is 1 g in 100 ml as it is mentioned as 1% w/V so it will be 10 g in 1000mL or 1 L
now get the moles of H2SO4 in 10 g  and convert it to molarity.Now you can use C1V1= C2V2

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