Dilution problem when only working with grams

  • #1
Homework Statement:
You have bottle with an unknown solution, you take 0.801g from it and put in a beaker, you add 43.883g of distilled water to the beaker. If you take 2.34g from the beaker, how many grams would that be if you had taken from the bottle instead with the unknown solution.
Relevant Equations:
C1V1=C2V2
Hi,

So I am aware of the C1V1=C2V2 equation but I feel that this is not really applicable here.

I attempted to go about it through instinct, so I would take 0.801g/43.883g = 0.0182. And then just multiply 0.0182 with 2.34g, to get 0.0426g.
So the answer would be, if I took 2.34 g from the beaker, I could have instead taken 0.0426g from the bottle with the unknown solution.

No idea if that is correct, what do you think?

But what also is bothering me is, if someone ask me, why not take 0.801/(43.883+0.801) as the ratio instead? I wouldn't know how to answer that..
Is there an equation I can refer to?

Thanks for your time
 

Answers and Replies

  • #2
Borek
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It is about mass fraction (which is a way of expressing concentration, but your intuition was right - it doesn't follow the simple dilution formula).

Mass fraction is a ratio of mass of the solute to the mass of the solution - like percent concentration, just without multiplying by 100. So if there is msubst grams of substance dissolved in msolv grams of solvent, mass fraction is

[tex]f = \frac {m_{subst}}{m_{subst} + m_{solv}}[/tex]

Assume initial (unknown) mass fraction to be x. Can you use information given to express new mass fraction (let's call it x') in terms of x? Can you use x' to express mass of the solute in 2.34 g of the new solution? If so, you are just a small step from getting the final answer using all the things you have calculated (and defined) so far.
 
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  • #3
Thank you Sir for your quick reply.

Sadly I did not understood what you meant. I need more clarification. However, I see that you asked in a way that, if I understood you, it would have been a very leading (good) question. So thanks for that.

But I realized in one way, that by taking everything in the beaker, I should have also taken the 0.801g I added.
Thing is, the amount in the beaker is not 43.883g, it is 43.833+0.801g.

So my original answer of 0,0426g is wrong, it should be 0,0419g.

I guess that is my way of reasoning around the question.
Thanks for the equation, mass fraction, I will remember that :smile:
 
  • #4
Borek
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So my original answer of 0,0426g is wrong, it should be 0,0419g.

Yes.

Hard to clarify not knowing what it was you didn't get.
 
  • #5
symbolipoint
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I did not yet read the posts which followed:

Problem StatementYou have bottle with an unknown solution, you take 0.801g from it and put in a beaker, you add 43.883g of distilled water to the beaker. If you take 2.34g from the beaker, how many grams would that be if you had taken from the bottle instead with the unknown solution.Relevant EquationsC1V1=C2V2

Your units in use are grams. Forget the volume!
 

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