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# How do you calculate osmolarity of a solution?

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#### Explanation

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#### Explanation:

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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.

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 1

Calculate the osmolarity of blood. The concentrations of solutes are:

["Na"⁺] = "0.140 mol/L";
$\text{[Glucose]" = "180 mg/100 mL}$;
$\text{[BUN] (blood urea nitrogen)" = "20 mg/100 mL}$.

Solution

$\text{[Na"^+"]" = "0.140 mol/L}$.

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

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

$\text{Glucose osmolarity" = (0.180 cancel("g"))/(100 cancel("mL")) × (1000 cancel("mL"))/"1 L" × (1 cancel("mol"))/(180.2 cancel("g")) × "1 Osmol"/(1 cancel("mol")) = "0.009 99 Osmol/L}$

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

$\text{Blood osmolarity" = "(0.280 + 0.009 99 + 0.0071) Osmol/L"= "0.297 Osmol/L" = "297 mOsmol/L}$

EXAMPLE 2

Calculate the osmolarity of an IV admixture that contains 500 mL sterile water; 50 mL NaHCO₃ 8.4 %; 10 mL of 2 mmol/mL KCl; 0.5 mL heparin 5000 units; 1 mL pyridoxine; 1 mL thiamine.

Solution

Set up a table for easy calculation.

$\text{Osmolarity" = "141.96 mOsmol"/(562.5 cancel("mL")) × (1000 cancel("mL"))/"1 L" = "252 mOsmol/L}$

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