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Chemistry, mL required to Concentration of HNO3

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    How many mL of 50.0% by mass HNO3 solution, with a density of 2.00 g/mL is required to make a 500 mL of a 2.00 M HNO3?


    2. Relevant equations

    initial Molarity X initial Volume = final Molarity X final Volume


    3. The attempt at a solution

    (50 g HNO3)(1 mL/2.00 g)(0.001 L/1 mL) = 0.025 L
    (50 g HNO3)(1 mol HNO3/63.012 g HNO3) = 0.79 mol

    0.79 mol/0.025 L = 31.6 M

    (500 mL)(2 M) = (31.6 M)V

    V = 31.6 mL


    Comments

    I don't know if the process that I used is right. Which leads me to doubt the answer. Any help is much appreciated.
     
  2. jcsd
  3. Dec 10, 2008 #2

    Borek

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    Staff: Mentor

    Generally not bad, although you did one serious mistake:

    That would be true if you would start with pure acid, not with the solution. 50 g of the solution has volume of 0.025L, that's OK, but it doesn't contain 50 g of HNO3.
     
  4. Dec 10, 2008 #3
    Thanks for the reply.

    Ok so the volume I obtained is right, but the label is wrong? And the answer is around 31.6 mL?
     
  5. Dec 10, 2008 #4

    Borek

    User Avatar

    Staff: Mentor

    No, this is a serious mistake (serious in terms of the way it changes final result).
     
  6. Dec 12, 2008 #5
    Ok ok, so since it says 50 percent by mass HNO3 I assumed that it was part of a 100 percent solution. Meaning 100 grams. So I did the whole process again with the new calculations and got 63.0 mL.

    Here is my work:

    (50 g solution)(1 mL/2.00 g)(0.001 L/1 mL) = 0.05 L
    (50 g HNO3)(1 mol HNO3/63.012 g HNO3) = 0.79 mol

    0.79 mol/0.05 L = 15.8 M

    (500 mL)(2 M) = (15.8 M)V

    V = 63.0 mL

    Which I found out later, from the answer key that my professor provided me with, was correct. Thanks!
     
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