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$$\pi V = nRT$$
What is the osmotic pressure at $\pu{25^oC}$ of an aqueous solution of 0.0010 M $\ce{C12H22O11}$ (sucrose)?
We just need to substitute the data into equation (13.4)

$$\pi = \frac{\pu{0.0010 mol} \times \pu{0.08206 atm L mol-1 K-1 }\times \pu{298 K}}{1L} \\ \pi = \pu{0.024 atm} (\pu{18 mmHg)}$$

I'm a bit confused about osmotic pressure.

I am not sure what $\pi$ represents in the example. Is this pressure the pressure needed to prevent the flow of solvent through a membrane into the solution? Why isn't it considering the amount of solvent outside? for example, if I had 1000L of outer water in a sucrose solution, wouldn't a different osmotic pressure be needed than if I only had 50L of outer water?

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    $\begingroup$ Is this pressure the pressure needed to prevent the flow of solvent through a membrane into the solution? - Yes it is. And it depends on the solvent volume just indirectly, if it causes hydrostatic pressure on the solvent side. $\endgroup$
    – Poutnik
    Commented Apr 25 at 17:53

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The pi (π) in the question represents the osmotic pressure, which is the pressure required to prevent the flow of solvent (water) through a semi-permeable membrane into the solution (higher concentration side). Osmotic pressure itself is independent of the solvent volume (outside) for ideal solutions at equilibrium.

For ideal solutions, the pressure needed to stop osmosis, or osmotic pressure, depends only on the sucrose concentration and NOT the outside water volume, whether that is 1000ML or 500ML. The larger water volume just means it takes longer to dilute the sucrose solution and reach equilibrium. It does NOT affect the osmotic pressure in any way.

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    $\begingroup$ Why would it take longer to dilute? $\endgroup$
    – Buck Thorn
    Commented Apr 26 at 7:17
  • $\begingroup$ The process might take longer to dilute because the larger volume of dilute water provides more water molecules available for diffusion into the concentrated solution. $\endgroup$
    – Ronith
    Commented Apr 26 at 11:51

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