[Ebbe Loos]

Hi

I need to find the concentration in g/L for AdBlue. I've found online that it is a 32,5% Urea concentration in demi water. Does this means that there is a 32,5g Urea in 100g of solution (water), so that the concentration in g/L is equal to 325g/L?

I'm having a hard time which makes me feel stupid, this seems like a easy question.

A friend of mine argues that since there is 32,5% Urea there needs to be 67,5% Water. But if you take these numbers you"ll get a concentration of 481,48 g/L.

Can somebody help me?

Thanks!

[billnotgatez]

from GOOGLE

A solution is a homogeneous mixture consisting of a solute dissolved into a solvent.
The solute is the substance that is being dissolved, while the solvent is the dissolving medium.

A percentage solution is an amount or volume of chemical or compound per 100 mL of a solution.

------------
water can not be both solvent and solution

[Ebbe Loos]

Thanks for the reply!

Concluding from your answer, i think that the answer would be 325g/L?

[billnotgatez]

how much water?

[Ebbe Loos]

What do you mean with 'how much water'??

I conclude that: if i were to make a 1L solution of 32,5% Urea in water. I need to add 325 grames of urea and add it to 1L of water.

Do you agree?

[billnotgatez]

C H E M I C A L   P E R C E N T   C A L C U L A T O R
Fill-in All Datafields:*
Percent desired (%): w/v 32.5%
Amount of solution needed: 1000 ml
The Answers:
Remember to start with less liquid than the amount needed. (about 650 ml)
then add Amount of chemical needed:   325 grams (g)
Add liquid to make a total volume of 1000 ml.

note red is my suggestion

Let us know how much water you had to add  to make the 1000 ml

[Borek]

A percentage solution is an amount or volume of chemical or compound per 100 mL of a solution.

No idea where you got it, but it is definitely not true. This is a lousy approach to prepare solutions which are later called w/v%, and it is one of the most ambiguous ways of expressing solution composition.

What do you mean with 'how much water'??

I conclude that: if i were to make a 1L solution of 32,5% Urea in water. I need to add 325 grames of urea and add it to 1L of water.

No, if you add 325g of urea to 1L of water you will end with 1.24L of solution that is 24.5%.

Percentage is [itex]\frac{mass~of~the~solute}{mass~of~the~solution}\times100\%[/itex], mixing 325 g of urea with 1L of water produces (more or less, depends on the temperature) 1325g of the solution.

[billnotgatez]

I guess all is not true on the internet below is the source

https://www.labce.com/spg913372_what_is_a_percentage_solution.aspx

A percentage solution is an amount or volume of chemical or compound per 100 mL of a solution.

No idea where you got it, but it is definitely not true. This is a lousy approach to prepare solutions which are later called w/v%, and it is one of the most ambiguous ways of expressing solution composition.
...

[billnotgatez]

Also from my post with
C H E M I C A L   P E R C E N T   C A L C U L A T O R
the source was
https://www.ou.edu/research/electron/bmz5364/calc-percent.html

[Ebbe Loos]

A percentage solution is an amount or volume of chemical or compound per 100 mL of a solution.

No idea where you got it, but it is definitely not true. This is a lousy approach to prepare solutions which are later called w/v%, and it is one of the most ambiguous ways of expressing solution composition.

What do you mean with 'how much water'??

I conclude that: if i were to make a 1L solution of 32,5% Urea in water. I need to add 325 grames of urea and add it to 1L of water.

No, if you add 325g of urea to 1L of water you will end with 1.24L of solution that is 24.5%.

Percentage is [itex]\frac{mass~of~the~solute}{mass~of~the~solution}\times100\%[/itex], mixing 325 g of urea with 1L of water produces (more or less, depends on the temperature) 1325g of the solution.

Thanks for the respons. I'm assuming you're talking about the w/w%. In this case: m_u/(m_u+m_w) = 32,5% and m_u+m_w = 1000g

If i solve this, I find that a solution of 325g Urea in 675g Water results in a 32,5 w/w%.
Resulting in a concentration of 481,48 g/L


I found this online: 'AdBlue is made by mixing urea with demineralized water resulting in a 32.5% aqueous urea solution.' (I'll provide link down below)

Do you have any idea of they mean w/w% or w/v%?

Anyways, i want to thank you both for this discussions.



link :https://www.vaisala.com/en/industries-applications/chemicals-allied-products/automotive-grade-urea-solution-process-adblue-and-def#:~:text=AdBlue%20is%20made%20by%20mixing,a%2032.5%25%20aqueous%20urea%20solution.

[Borek]

If i solve this, I find that a solution of 325g Urea in 675g Water results in a 32,5 w/w%.
Resulting in a concentration of 481,48 g/L

You are right about using 675 grams of water, but you are wrong about the g/L concentration. Volume changes, so it is not 0.675 L . The only way to find out the final volume is to either measure it, or use experimental density of the solution of a given concentration found in data tables (1.0883 g/mL for 32.5% urea).

Do you have any idea of they mean w/w% or w/v%?

Sorry, no. But the only one that is unambiguous is w/w so I would assume that's what they mean.

[Ebbe Loos]

You are right about using 675 grams of water, but you are wrong about the g/L concentration. Volume changes, so it is not 0.675 L . The only way to find out the final volume is to either measure it, or use experimental density of the solution of a given concentration found in data tables (1.0883 g/mL for 32.5% urea).

I need to find the answer to the question: how much gram of Urea do you need to add to 1L off water to get a 32,5% solution. So like the methode of previous post:

Add 481,48g of Urea to 1000g of water to get a solution of 32,5 w/w%.

But why can't i say that the concentration of Urea in the water is equal to 481,48g/L? Since i dissolve the urea in water?

Anyways using the density: total solution mass = 1481,48g this solves to a volume of 1367,52 mL.
Is the concentration now of Urea in the solution now equal to: 481,48g / 1367,52 mL = 352,08 g/L?

But now i have 2 concentrations. You see my problem?

I want the concentration since it's easier to work with; since you can just read that mass you need to add to 1L of water

Hope you can help me

[Borek]

But why can't i say that the concentration of Urea in the water is equal to 481,48g/L? Since i dissolve the urea in water?

Because the volume changes - when you add a lot of urea to 1L of water the volume of the solution is no longer 1L.

And note that when someone says "x w/v%" they sometimes mean "xg in 100mL of the solution" and sometimes "xg dissolved in 100mL of solvent", which adds further ambiguity. Avoiding w/v is the only sure way of avoiding problems.

Anyways using the density: total solution mass = 1481,48g this solves to a volume of 1367,52 mL.
Is the concentration now of Urea in the solution now equal to: 481,48g / 1367,52 mL = 352,08 g/L?

Yes.

But now i have 2 concentrations.

No, you don't have two concentrations - you have a concentration and a random, incorrect number which you mistakenly call concentration

w/w is unambiguous, because masses are additive - xg of water plus yg of urea produces (x+y)g of mixture. When you mix volumes of different liquids, or when you add a solid to the liquid, final volume is very rarely equal to sum of volumes (and if it is, it is by a happy accident, because the general rule is "volumes are not additive").