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

[Neophyte]

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 :yuck:

 

[Bohrok]

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

 

[Neophyte]

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.

 

[Bohrok]

Do you have the answer to the problem?

 

[Neophyte]

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?

 

[Bohrok]

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

 

[Borek]

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

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

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