Type 2: Find the weight of the solute from mOsmol and volume
5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General
5.2.1.1) mOsmol - General Normal 1
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If potassium citrate monohydrated solution is expressed as 0.1852 mOsmol/mL, then how many milligrams of potassium citrate monohydrated will be present in each 100 mL of the solution? The molecular weight of potassium citrate monohydrated is 324 g/mol.
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lb equals 1500 mg/100mL kg
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For potassium citrate monohydrated , the 'golden formula' is:
1 mmol = `[324]` mg = `[3]` mEq = `[4]` mOsmol.
Since we need mg in the numerator of the desired unit, we begin with keeping it in the numerator of the first fraction.
Again, since 0.1852 mOsmol is contained in 1 mL of solution, we keep the 0.1852 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
Using dimensional analysis:
`(324 \quad mg)/(4 \quad mOsmol) × (0.1852 \quad mOsmol)/(1 \quad mL) × (100 \quad mL)/(100 \quad mL)= 1500 \quad (mg)/(100 \quad mL)` Ans.