How can I calculate osmolarity of blood?

May 10, 2014

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

Explanation:

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

For example, NaCl dissociates completely in water to form Na⁺ ions and Cl⁻ ions. Thus, each mole of NaCl becomes two osmoles in solution: one mole of Na⁺ and one mole of Cl⁻.

A solution of 1 mol/L NaCl has an osmolarity of 2 Osm/L.

A solution of 1 mol/L CaCl₂ has an osmolarity of 3 Osm/L (1 mol Ca²⁺ and 2 mol Cl⁻).

EXAMPLE

Calculate the osmolarity of blood. The concentrations of solutes are: [Na⁺] = 0.140 mol/L; [glucose] = 180 mg/100 mL; [BUN] (blood urea nitrogen) = 20 mg/100 mL.

Solution

[Na⁺] = 0.140 mol/L. But, each Na⁺ ion pairs with a negative ion X⁻ such as Cl⁻ to give 2 Osm of particles.

∴ NaX osmolarity = $\left(0.140 \text{ mol")/(1" L") × (2" Osm")/(1" mol}\right)$ = 0.280 Osm/L

Glucose osmolarity =

$\left(0.150 \text{ g")/(100" mL") × (1000" mL")/(1" L") × (1" mol")/("180.2 g") × (1" Osm")/(1" mol}\right)$ = 0.008 32 Osm/L

BUN osmolarity = $\left(0.020 \text{ g")/(100" mL") × (1000" mL")/(1" L") × (1" mol")/(28.01" g") ×(1" Osm")/("1 mol}\right)$ = 0.0071 mol/L

∴ Blood osmolarity = (0.280 + 0.008 32 + 0.0071) Osm/L = 0.295 Osm/L = 295 mOsm/L