Type 2: Find the weight of the solute from mOsmol and volume

5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General 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

    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.