Type 3: Determine percent strength from osmolarity of the solution

5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General mOsmol - General Normal 1

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The osmolarity of a calcium gluconate solution is 115.12 mOsmol/L. How will this be expressed as % w/v strength? Given that, the molecular weight of calcium gluconate is 430 g/mol.

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lb equals 1.65 % kg

    For calcium gluconate , the 'golden formula' is:
    1 mmol = `[430]` mg = `[2]` mEq = `[3]` mOsmol.

    Osmolarity is the number of mOsmol per 1 L (or 1000 mL).
    Since we need g quantity in the numerator of the desired unit, we begin with keeping the mg unit in the numerator of the first fraction.
    Again, since 115.12 mOsmol is contained in 1000 mL of solution, we keep the 115.12 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
    Using dimensional analysis:

    `(430 \quad mg)/(3 \quad mOsmol) × (115.12 \quad mOsmol)/(1000 \quad mL) × (1 \quad g)/(1000 \quad mg)`

    ` × (100 \quad mL)/(100 \quad mL)= 1.65 \quad (g)/(100 \quad mL) => 1.65 % ` Ans.