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 5 mL of dextrose 5% solution in half-normal saline (D5½NS).
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lb equals 2.16 mOsmol kg
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Dextrose 5% in half-normal saline means 5g of dextrose and 0.45 g of NaCl is present per 100 mL of the solution. The volume of the solution is 5 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) × 5 \quad mL`
`= 1.39` mOsmol.
NaCl is dissociated to two particles; therefore, we can write: `[58.5]` mg = `[2]` mOsmol.
Using dimensional analysis, mOsmol contributed from NaCl:
`(0.45 \quad g)/(100 \quad mL) × (1000\quad mg)/(1\quad g) × (2\quad mOsmol)/(58.5\quad mg) × 5 \quad mL`
`= 0.77` mOsmol.
Total mOsmol will be the sum of these two. Therefore, the mOsmol of the solution is:
`1.39 \quad mOsmol + 0.77 \quad mOsmol = 2.16 \quad mOsmol`. Ans.