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Homework Help: Beginner analytical chemistry question

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    you want to prepare 500 ml of 1 M KNO3 at 20 C, but your lab and water are 24 C. How many grams of solid KNO3 (density= 2.109 g/ml) should be dissolved in a volume of 500 ml at 24 C so that the concentration will be 1 M at 20 C? what apparent mass of KNO3 weighed in air is required?

    2. Relevant equations

    Molarity = moles solute/liters solvent

    and dimensional analysis


    3. The attempt at a solution

    i'm really not sure what to do with this..typically i am good with this kind of question but it was kind of thrown at us, and i've never seen the temperature issue before. please help!
     
  2. jcsd
  3. Jan 27, 2010 #2

    chemisttree

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    Is that really the question? 1 M KNO3? Not 1.000 M?

    Your solvent at 24C will occupy a larger volume than at 20C so there is one source of error. That said, your vessel (volumetric flask) will also change its dimesions at 24C vs at 20C that will counteract this tendency. Unless you know both the dV of the vessel and the dV of the solution over the 4C range, you don't know how to solve it analytically. But I digress...

    Did the question really ask for 1 M KNO3 or 1.000 M KNO3?

    And... will you use the solution at 20C or 24C?
     
  4. Jan 27, 2010 #3
    it said 1.000 M KNO3. sorry for not specifying.. i was really tired when i typed it out.

    i don't know either of these things that you mention. This is only my 2nd day in analytical chem, and this question was given to me as an assignment.

    i also think the question is unclear.. but, obviously, there's nothing we can do about it

    thanks for your help anyway!
     
  5. Feb 9, 2010 #4
    i actually solved this problem

    the key lies in a formula that was quite difficult to find at first, but quite simple to use afterwards

    c1/d1 = c2/d2, where the different concentrations and densities are due to the temperature difference

    after that, the buoyancy correction equation is used

    thanks for your help anyway
     
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