How do you calculate osmolarity of a solution?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

199

Answer:

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, #"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 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.

enter image source here

#"Osmolarity" = "141.96 mOsmol"/(562.5 cancel("mL")) × (1000 cancel("mL"))/"1 L" = "252 mOsmol/L"#

Was this helpful? Let the contributor know!
1500