# How do you calculate osmolarity from molarity?

Mar 20, 2014

You multiply the molarity by the number of osmoles that each solute produces.

#### Explanation:

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

For example, $\text{NaCl}$ dissociates completely in water to form ${\text{Na}}^{+}$ ions and $\text{Cl"^"-}$ ions.

Thus, each mole of $\text{NaCl}$ becomes two osmoles in solution: one mole of ${\text{Na}}^{+}$ and one mole of $\text{Cl"^"-}$.

A solution of 1 mol/L $\text{NaCl}$ has an osmolarity of 2 Osmol/L.

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

EXAMPLE

Calculate the osmolarity of blood.

Solution

The concentrations of solutes are: $\text{[Na"^+] = "0.140 mol/L}$; ["glucose"] = "180 mg/100 mL"; $\text{[BUN] (blood urea nitrogen)" = "20 mg/100 mL}$.

["Na"^+] = "0.140 mol/L".

But, each ${\text{Na}}^{+}$ ion pairs with a negative ion $\text{X"^"-}$ such as $\text{Cl"^"-}$ to give 2 Osmol of particles.

$\text{NaX osmolarity" = (0.140 color(red)(cancel(color(black)("mol"))))/"1 L" × "2 Osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.280 Osmol/L}$

$\text{Glucose osmolarity" = (0.150 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("mL")))) × (1000 color(red)(cancel(color(black)("mL"))))/"1 L" × (1 color(red)(cancel(color(black)("mol"))))/(180.2 color(red)(cancel(color(black)("g")))) × "1 Osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.008 32 Osmol/L}$

$\text{BUN osmolarity" = (0.020 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("mL")))) × (1000 color(red)(cancel(color(black)("mL"))))/"1 L" × (1 color(red)(cancel(color(black)("mol"))))/(28.01 color(red)(cancel(color(black)("g")))) ×"1 Osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.0071 Osmol/L}$

$\text{Blood osmolarity" = "(0.280 + 0.008 32 + 0.0071) Osmol/L" = "0.295 Osmol/L" = "295 mOsmol/L}$