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Homework Help: Moles of solute particles are present in 5.07 mL of 0.688 M

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    How many moles of solute particles are present in 5.07 mL of 0.688 M Na3PO4? Use scientific notation with 3 significant figures!!!!

    2. Relevant equations

    3. The attempt at a solution
    0.00507 L of soln
    0.688 M
    M= mol solute / L of soln
    0.688 = x / 0.00507
    .688 * 0.00507
    x = 3.49E-3

    What went wrong :yuck:
  2. jcsd
  3. Sep 30, 2009 #2
    Re: Molarity

    Moles of solute particles should be osmolarity: the molarity of aall particles which would be the molarities of the two ions.
  4. Sep 30, 2009 #3
    Re: Molarity

    So it is
    3.49E-3 * 4 = 1.40E-2 ?
    Thank you for your help.
  5. Sep 30, 2009 #4
    Re: Molarity

    Do you have the answer to the problem?
  6. Sep 30, 2009 #5
    Re: Molarity

    No, it was simply marked incorrect so I am unsure of the actual answer.

    I just looked up osmolarity after you mentioned it. My book unfortunately neglected it :(. This is the first encounter with it so I was not sure if the method of obtaining it was correct. "the osmolarity of a simple solution is equal to the molarity times the number of particles per molecule"

    I had the molarity = 3.49E-3
    (Na3 = 3 PO4 = 1) = 4
    3.49E-3 * 4

    If it said "How many moles of solute are present" would it have been just finding the molarity?
  7. Sep 30, 2009 #6
    Re: Molarity

    Yes, "moles of solute" would be molarity, but moles of solute particles would mean osmolarity.
  8. Oct 1, 2009 #7


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    Staff: Mentor

    Re: Molarity

    You may treat it as a nitpicking (and I won't comment) but this is not the correct result :wink:

    I mean - on some level it works, but phosphate is a strong base, it reacts with water and Na+ to some extent reacts with OH-. Thus the real result is slightly different - total concentration of dissolved entities (be it ions or neutral molecules) is about 2.804M, and number of moles is 14.2*10-3 (not 14.0*10-3).

    Even that is only an approximation, but better than the initial one :smile:


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