Type 4: Osmolarity of IV solutions (D5W, D5NS, D5½NS)

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

5.2.1.1) mOsmol - General Hard 1


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Calculate the total milliosmoles of 15 mL of dextrose 5% solution in normal saline (D5NS).


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lb equals 8.79 mOsmol kg

    Dextrose 5% in normal saline means 5g of dextrose and 0.9 g of NaCl is present per 100 mL of the solution. The volume of the solution is 15 mL.
    Dextrose is not dissociated in the solution, therefore, we can write the 'golden formula' as:
    `1\quad mmol= [180]` mg = `[1]` mOsmol.

    Using dimensional analysis, mOsmol contributed from dextrose:

    `(5 \quad g)/(100 \quad mL) × (1000\quad mg)/(1\quad g) × (1\quad mOsmol)/(180\quad mg) × 15 \quad mL`
    `= 4.17` mOsmol.

    NaCl is dissociated to two particles; therefore, we can write: `[58.5]` mg = `[2]` mOsmol.

    Using dimensional analysis, mOsmol contributed from NaCl:

    `(0.9 \quad g)/(100 \quad mL) × (1000\quad mg)/(1\quad g) × (2\quad mOsmol)/(58.5\quad mg) × 15 \quad mL`
    `= 4.62` mOsmol.

    Total mOsmol will be sum of these two. Therefore, the mOsmol of the solution is:
    `4.17 \quad mOsmol + 4.62 \quad mOsmol = 8.79 \quad mOsmol`. Ans.