Type 1: Determine mOsmol from weight or percent of the solute
5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General
5.2.1.1) mOsmol - General Normal 1
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A solution contains 270 mg of potassium citrate monohydrated in each 15 mL (the MW of potassium citrate monohydrated is 324). Calculate the osmolarity of this solution
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lb equals 222.22 mOsmol/L kg
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For potassium citrate monohydrated , the 'golden formula' is:
1 mmol = `[324]` mg = `[3]` mEq = `[4]` mOsmol.
Osmolarity is the number of mOsmol in 1000 mL (1 L) of the solution.
Using dimensional analysis:
`(270 \quad mg)/(15 \quad mL) × (3.3333 \quad mOsmol)/(324 \quad mg) × (1000 \quad mL)/(1 \quad L) = 222.22 \quad (mOsmol)/(1 \quad L)`.
Therefore, the osmolarity is 222.22. Ans.