I have a concentration of saltwater (NaCI) in that there was 23 g of salt dissolved in .025 L of water.

Question

How many atoms were dissolved?

Is this how you set it up???
.39 x(6.022x10^23)

Answer 1

I think that will give you the number of "molecules" of NaCl. I would think twice that would be the number of atoms dissolved if we call the ions in solution atoms. Technically there were no "atoms" dissolved, only ions.

Answer 2

I am a little confused ... the previous questions were
What is the concentration in W/V units?
(23g/250ml)x100 = 9.2% w/v

How many moles of NaCi were dissolved?

(23g/1) x (1mol/58.54g)+ .39moles

Then it was
How many atoms were dissolved?
and this is zero because there are no atoms only ions??????

Answer 3

I agree with the 9.2% w/v as well as the number of mols NaCl. I assume this is a beginning chemistry course and I also assume that the author of the problem just "made a mental error" in writing the problem and PROBABLY meant to ask how many ions were dissolved which will be twice the number of mols and that times 6.02E23. As I posted earlier, however, there are no atoms dissolved. Even the solid NaCl crystal has no atoms because the solid NaCl crystal is composed of sodium ions and chloride ions. In solution, the crystal lattice bonds have been broken and the Na^+ ions and Cl^- ions have become solvated and you have floating around Na^+(H2O)n and Cl^-(H2O)n

Answer 4

Ohhhh Ok thank you sooo much :) makes more sense Thank you again