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When people talk about osmosis, they say that when we add a mole of NaCl, then each its molecule gets dissolved into 2 separate particles and therefore we have 2osM/L solution.

But both Na & Cl are attached to molecules of water which means that those water molecules are not water anymore and they can't penetrate the membrane. Thus not only the presence of Na & Cl itself changes the concentration of water in solution, but also the number of water molecules taken by those particles.

Therefore if we put 1 particle that binds to 1 water molecule and 1 particle that binds to 5 water molecules - they change the water concentration differently.

Please somebody tell me where I'm wrong.

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Just wanted to add that the kind of salt or solvent in the water does matter. The Chemical Activity of the ions or other soluates matter quite a bit. For example osmosis can be caused by sugar, organic molecules fats as well as salts. Here, Chemical Activity is a general idea of the relative effect of different molecules on osmosis.

Different soluates can cause drastically different effects on osmosis. e.g. see Table 1 in this reference.

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You can see the importance of the number of molecules if you imagine a semipermeable membrane, on one side there is water, and on the other there is the salty water solution. Water molecules can pass the membrane but ions cannot. On the side of the solution, because Na and Cl are taking space and covering some parts of the membrane, the water molecules are "touching" the membrane less, so the flow of water molecules out from solution side is slower than the pure water side. This means that more water is flowing into the solution than out of it, slowly decreasing the concentration of the solution, until they reach an equilibrium when the flow of water to both sides is equal.

You are incorrect to say that the "water molecules are no longer water", because their interactions with the ions are intermolecular (ion-dipole), not intramolecular. Intermolecular interactions may slow down molecules, but since they do not lock the water molecules in place, they could still pass the membrane. Although, the concentration of solute is important in calculating osmotic pressure. The formula used when the contentration is low is different than if the concentration was high. For the formuals see Osmotic Pressure on Wikipedia.

The visualization with the membrane is my interpretation of Silberberg's explanation of osmotic pressure in Principles of Chemistry, page 412.

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When you dissolve something in water, you build a solvation shell around this - so each ion (in the case of salts) gets surrounded by a number of water molecules. For more details read this Wikipedia article. The concentration change in free water molecules is usually not taken into account since it can be neglected due to the vast number of "unbound" molecules. So if you have one litre of water, it contains approximately 56 moles of water molecules. If you dissolve 1 mole of sodium chloride in this water, and take in account that this takes quite some water to get solvated, there is ways more water available. So the water concentration is almost constant (especially when you work at physiological osmosis condition).

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  • $\begingroup$ Okay, let's calculate this scenario: we add 1 mole of NaCl (~60g) into 56 moles of water. Each Cl atom binds to 6 atoms of water at average (according to wiki: en.wikipedia.org/wiki/Sodium_chloride). Let's suppose that Na takes up the same number of water molecules. Then 1 mole of salt takes 12 moles of water which doesn't look negligible to me. On practice the number will probably be lower because a single water molecule can be bound to both Na & Cl atoms which means that in highly concentrated solution 1 mole of NaCl will bind to 6 moles of water which is still not negligible. $\endgroup$ Feb 5, 2014 at 7:56
  • $\begingroup$ It indeed sounds that its not neglectable. But you still have 80% of the water present. And: Water is a neutral molecule that can pass through the cell membrane, other dissolved substances cannot do this. $\endgroup$
    – Chris
    Feb 5, 2014 at 10:25

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