Understanding the Osmotic Pressure formula

Question

$$\Pi = iMRT$$

Can this be rewritten as $\Pi = ORT$ where $O =$ osmolarity? The only reason I ask is because my book doesn’t talk about the relationship between osmolarity and molarity, but I came across a practice problem where I had to find the minimum pressure required to purify sea water given the osmolarity of the seawater and temperature (no compound was given).

Also, this practice problem got me thinking about the relationship between osmolarity and molarity. Without knowing what the compound is in the seawater, there’s no way to determine the molarity, correct?

A compound like glucose would have a van’t Hoff factor of 1 so it’s molarity would be equal in value to the osmolarity. However, a compound like $\ce{NaCl}$ would have a molarity equal to half the osmolarity.

Here is the word problem for context:

Reverse osmosis is a process that allows fresh water to be obtained by using pressure to force an impure water source through a semi-permeable membrane that only allows water molecules to pass. What is the minimum pressure that would be required to purify seawater at $\pu{25 ^\circ C}$ that has a total osmolarity of $\pu{1,000 mOsm L-1}$?

I don’t want an answer to this problem because I have the answer and I know how to calculate it. But if it helps to answer the general question, please feel free to use the problem as a framework for a more general answer.

Answer 1

Your formula is correct. You can always calculate the molarity of a solution if you know the amount of substance dissolved in a lieter of solution and also you know the formula of the solute. For instance, if you dissolve $\pu{1 mol}$ of sugar, or of alcohol, or of glycerol in water and the solution made upto $\pu{1 L}$, you obtain a solution which is $1$ molar $(\pu{1 M})$, and also $1$ osmolar $(\pu{1 Osm})$, because these substances do not dissociate in water (no ions).

If you dissolve $\pu{1 mol}$ $\ce{ NaCl}$ $(\pu{58.5 g mol-1})$, or $\pu{1 mol}$ $\ce{KNO3}$ $(\pu{101 g mol-1})$ in $\pu{1 L}$ solution, you obtain a solution which is $1$ molar, but $2$ osmolar, because each of $\ce{NaCl}$ and $\ce{KNO3}$ produce two ions in the solution. The osmolarity O is $2$ osmol/L in such solutions.

If you dissolve $\pu{1 mol}$ $\ce{CaCl2}$ $(\pu{111 g mol-1})$ or $\pu{1 mol}$ $\ce{Ba(OH)2}$ in $1$ liter water, you obtain a solution which is $1$ molar and $3$ osmolar, because each of $\ce{CaCl2}$ or $\ce{Ba(OH)2}$ produce three ions in solution. The osmolarity O in such a solution is $3$ osmol/L. This calculation is rather theoretical, because it it not possible to dissolve such a huge amount as $\pu{1 mol}$ $\ce{Ba(OH)2}$ in one liter water.