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22/09/2021
You multiply the by the number of osmoles that each solute produces.
An osmole (Osmol) 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 of 2 Osmol/L.
A solution of 1 mol/L ##”CaCl”_2## has an osmolarity of 3 Osmol/L (1 mol ##”Ca”^(2+)## and 2 mol ##”Cl”^-##).
EXAMPLE 1
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 Osmol of particles.
##”NaX osmolarity” = (0.140cancel(“mol”))/(1″L”) × “2 Osmol”/(1cancel(“mol”)) = “0.280 Osmol/L”##
##”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”##
##”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”##
##”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.
##”Osmolarity” = “141.96 mOsmol”/(562.5 cancel(“mL”)) × (1000 cancel(“mL”))/”1 L” = “252 mOsmol/L”##
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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