How is osmolarity related to molarity?

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Answer 1 · 2014-08-03T21:02:14

Osmolarity equals molarity times the van't Hoff $i$ $i$ factor.

Osmolarity is the number of osmoles of solute per litre of solution .

An osmole is 1 mol of particles that contribute to the osmotic pressure of a solution.

The term comes from the formula for calculating osmotic pressure:

$Π = M R T$ $Π = MRT$ , where $M$ $M$ is the molarity and $T$ $T$ is the Kelvin temperature.

Osmotic pressure is a colligative property. It depends on the number of particles.

If the solute is a nonelectrolyte like glucose, 1 mol of solute forms 1 mol of solute particles.

If the solute is an electrolyte, it will dissociate into ions.

$NaCl(s) \to {Na}^{+} (aq) + {Cl}^{-} (aq)$ $"NaCl(s)" → "Na"^+("aq") + "Cl"^(-)("aq")$

Thus, 1 mol of $NaCl$ $"NaCl"$ forms 2 mol of solute particles. It contains 2 Osmol of particles.

The osmotic pressure of a 1 mol/L $NaCl$ $"NaCl"$ solution will be twice that of a 1 mol/L solution of glucose.

The osmotic pressure formula becomes

$Π = i M R T$ $Π = iMRT$ , where $i$ $i$ is the van't Hoff factor.

The van't Hoff factor is the number of moles of particles formed from 1 mol of solute.

Thus, a solution of 1 mol/L $NaCl$ $"NaCl"$ has an osmolarity of 2 Osmol/L,

and a solution of 1 mol/L ${CaCl}_{2}$ $"CaCl"_2$ has an osmolarity of 3 Osmol/L (1 mol ${Ca}^{2 +}$ $"Ca"^(2+)$ and 2 mol ${Cl}^{-}$ $"Cl"^-$ .

We can rewrite the formula for osmotic pressure as

$Π = O s M \times R T$ $Π = OsM×RT$

If we compare the two formulas, we see that

$O s M = i \times M$ $OsM = i×M$

That is, the osmolarity is the molarity times the van't Hoff factor.

The video below explains the difference between molarity and osmolarity.