Dilution problem when only working with grams

In summary, if you took 2.34g from the beaker, you would have instead taken 0.0419g from the bottle with the unknown solution.
  • #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
 
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  • #2
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
Idontknowanything said:
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
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!
 

1. How do I calculate the dilution factor when working with grams?

The dilution factor can be calculated by dividing the final volume of the solution by the initial volume. For example, if you have 100 grams of a substance and you want to dilute it to a final volume of 500 mL, the dilution factor would be 500 mL / 100 mL = 5.

2. Can I use a balance to measure the initial and final masses in a dilution problem?

Yes, a balance can be used to measure the initial and final masses. However, it is important to make sure that the balance is calibrated and accurate to ensure precise measurements.

3. How do I determine the concentration of the diluted solution in grams per liter?

To determine the concentration of the diluted solution in grams per liter, divide the mass of the solute (in grams) by the volume of the solution (in liters). For example, if you have 50 grams of a solute in 500 mL of solution, the concentration would be 50 grams / 0.5 liters = 100 grams per liter.

4. Are there any common mistakes to avoid when working with grams in a dilution problem?

One common mistake to avoid is using the wrong units when calculating the dilution factor. Make sure to use consistent units of measurement (e.g. grams and milliliters) throughout the problem. It is also important to accurately measure the initial and final masses, as small errors can greatly affect the final concentration of the solution.

5. Can the dilution factor ever be greater than 1?

Yes, the dilution factor can be greater than 1. This would occur if you are diluting a solution with a larger volume of solvent than the initial volume of the solution. For example, if you have 100 grams of a solute in 50 mL of solution, and you dilute it to a final volume of 100 mL, the dilution factor would be 100 mL / 50 mL = 2.

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