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Dilution Calculation

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    150g of a solution which is 20% sulphuric acid (by mass) are added to x grams of a solution which is 70% sulphuric acid (by mass). What must be the value of x so that the final mixture shall contain 30% sulphuric acid (by mass)?

    2. Relevant equations

    a/b=c/d ? Not sure...this is my problem.

    3. The attempt at a solution

    I've tried equivalency calculation and any other ratio calculation that I could think of...all to no avail.

    I know that 30g of the original solution is SA and 120g is water. So know I need to raise that concentration (mass SA to mass Water) by 10% with a 70% solution. Where do I get started?
     
  2. jcsd
  3. Mar 21, 2009 #2
    I would use C1V1 = C2V2

    Where C = concentration in mols/L , and V = volume in Liters.
     
  4. Mar 21, 2009 #3
    I don't know mols or volume. I think you can solve this question (somehow) without that knowledge.
     
  5. Mar 21, 2009 #4
    Well the mass of the sulphuric acid for the first solution is (150 * 0.20)g wouldn't it?

    So from there you can solve for mols.

    Water has a density of 1g/1mL, so you have a volume.
     
  6. Mar 21, 2009 #5

    symbolipoint

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    This is nothing more than percentage and proportion. Your units are all by mass. (or grams, in this case).
     
  7. Mar 21, 2009 #6

    Borek

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    Staff: Mentor

    No need for moles, and C1V1=C2V2 is not the correct approach.

    All you need to use is definition of mass pecentage and mass conservation - whatever you add, it stays in the mix. If you mix two solutions, mass of water in the final solution will equal sum of masses of water in the input solutions. Same about sulfuric acid.
     
  8. Mar 21, 2009 #7
    Can you give me an example to get me started. It is probably way easier than it looks (to me).
     
  9. Mar 21, 2009 #8

    Borek

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    Imagine you mix 100g 50% sulfuric acid with 50 g 20% sulfuric acid.

    Using percentage definition, you get:

    first solution: 50 g acid, 50 g water
    second solution: 10 g acid, 40 g water
    ------------------------------------
    mixture: 60 g acid, 90 g water

    So the final solution is 40%.

    You have to somehow reverse the problem, but the general approach will be based on the same principles.
     
  10. Mar 21, 2009 #9
    Yeah I didn't read you question properly, C1V1= C2V2 will definitely not work.
     
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