Moles of solute particles are present in 5.07 mL of 0.688 M

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Discussion Overview

The discussion revolves around calculating the number of moles of solute particles in a solution of Na3PO4, specifically focusing on the distinction between moles of solute and moles of solute particles, which involves osmolarity. The context is primarily homework-related, with participants exploring the implications of molarity versus osmolarity.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the moles of solute using molarity and finds 3.49E-3 moles for 5.07 mL of 0.688 M Na3PO4.
  • Another participant suggests that the calculation should consider osmolarity, which accounts for the total number of particles in solution, proposing that the moles of solute particles should be 1.40E-2.
  • A participant expresses uncertainty about the correct approach, noting that their textbook did not cover osmolarity, and questions whether finding moles of solute would simply involve using molarity.
  • Another participant clarifies that "moles of solute" refers to molarity, while "moles of solute particles" refers to osmolarity.
  • One participant challenges the earlier calculations, suggesting that the actual concentration of dissolved entities is approximately 2.804 M due to the behavior of phosphate in solution, leading to a different estimate of moles.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method for calculating the moles of solute particles, with multiple competing views on the interpretation of osmolarity and the effects of chemical behavior in solution.

Contextual Notes

There are unresolved assumptions regarding the behavior of ions in solution and the definitions of molarity versus osmolarity, which may affect the calculations presented.

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Homework Statement



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

Homework Equations





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
 
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Moles of solute particles should be osmolarity: the molarity of aall particles which would be the molarities of the two ions.
 


Bohrok said:
Moles of solute particles should be osmolarity: the molarity of aall particles which would be the molarities of the two ions.

So it is
3.49E-3 * 4 = 1.40E-2 ?
Thank you for your help.
 


Do you have the answer to the problem?
 


Bohrok said:
Do you have the answer to the problem?

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?
 


Yes, "moles of solute" would be molarity, but moles of solute particles would mean osmolarity.
 


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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