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22/09/2021
You multiply the molarity by the number of osmoles that each 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.
Also, a solution of 1 mol/L ##”CaCl”_2## has an osmolarity of 3 Osmol/L (1 mol ##”Ca”^(2+)## and 2 mol ##”Cl”^”-“##).
EXAMPLE
Calculate the osmolarity of blood.
Solution
The concentrations of solutes are: ##”[Na”^+] = “0.140 mol/L”##; ##[“glucose”] = “180 mg/100 mL”##; ##”[BUN] (blood urea nitrogen)” = “20 mg/100 mL”##.
##[“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” = (0.140 color(red)(cancel(color(black)(“mol”))))/”1 L” × “2 Osmol”/(1 color(red)(cancel(color(black)(“mol”)))) = “0.280 Osmol/L”##
##”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”##
##”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”##
##”Blood osmolarity” = “(0.280 + 0.008 32 + 0.0071) Osmol/L” = “0.295 Osmol/L” = “295 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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