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 penicillin V potassium solution is expressed as 0.0309 mOsmol/mL, then how many milligrams of penicillin V potassium will be present in each 10 mL of the solution? The molecular weight of penicillin V potassium is 388 g/mol.

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lb equals 60 mg/10mL kg

    For penicillin V potassium , the 'golden formula' is:
    1 mmol = `[388]` mg = `[1]` mEq = `[2]` 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 0.0309 mOsmol is contained in 1 mL of solution, we keep the 0.0309 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
    Using dimensional analysis:
    `(388 \quad mg)/(2 \quad mOsmol) × (0.0309 \quad mOsmol)/(1 \quad mL) × (10 \quad mL)/(10 \quad mL)= 60 \quad (mg)/(10 \quad mL)` Ans.

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