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Simple molar ration problem

  1. Dec 29, 2015 #1
    I have been given NiAc and ZnAc solutions of 1M concentration. And being asked to mix the two solutions together in Ni/Zn nominal molar proportions 1%, 3% and 5%. How can I calculate and mix the two solutions to get the required molar proportion.
     
  2. jcsd
  3. Dec 29, 2015 #2

    Bystander

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    Ni/Zn = 0.01, 0.03, and 0.05. Calculate it and do it.
     
  4. Dec 30, 2015 #3
    Molar fraction = (number of moles of component of interest) ÷ (total number of moles of mixture)

    For Ni/Zn nominal proportion 1%
    0.01= number of moles of Ni/ total number of moles of Ni+Zn
    Suppose we have total moles of (Ni+Zn) solution= 5 mmol
    0.01= moles of Ni/ 5mmol
    Moles of Ni = 0.05 mmol
    Moles of Zn= total moles of (Ni+Zn)- moles of Ni
    Moles of Zn= 4.95 moles
    As both NiAc and ZnAc concentration is 1M.
    M=n/V
    V (ZnAc solution) = 4.95mmole/1 mol/l = 4.95ml

    V (NiAc solution) = 0.05mmole/1 mol/l = 0.05ml

    is it correct calculation?
     
    Last edited: Dec 30, 2015
  5. Dec 30, 2015 #4

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    That's one. Two to go.
     
  6. Dec 30, 2015 #5

    epenguin

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    No it isn't, though the error is small. You have made a solution which is 0.01 M Ni, but that is not what was asked for. It is not quite 1M Zn any more, so 1 % is not exactly the molar proportion.
    If the starting molarities are equal an you are asked for a 1 % ratio you just mix them in the ratio 100:1.
    You're probably so used to problems of achieving a given final molarity that you have overthught this one
     
  7. Dec 30, 2015 #6

    Bystander

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    Divalent ions, one to one.
     
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