Question

How many mOsmol/L are in 5% Dextrose in 0.45% NaCl?

Answer 1

5% Dextrose means 5g of dextrose are dissolved in 100mL of water similarly 0.45% of NaCl means 0.45g of NaCl are dissolved in 100mL of water.

No. of moles of NaCl in 100mL = mass of NaCl / molar mass of NaCl = 0.45 / 58.5 = 0.0077 moles

Hence moles of NaCl in 1000mL (or 1L) = (0.0077 X 1000) / 100 = 0.077 moles = 77mmoles

Similarly moles of dextrose in 100mL = Mass of Dextrose / Molar mass of dextrose = 5 / 180.15 = 0.0277 moles

Moles of dextrose in 1L = (0.0277 X 1000) / 100 = 0.277 moles = 277 mmoles

No. of osmoles = number of moes of particles produced

1 mmole of NaCl produces 1 mmole of Na⁺ and 1 mmol of Cl^-.

Or 1 mmole of NaCl produces 2 mOmoles of particles

Therefore number of mOsmoles produced by 77 mmole of NaCl = 77 mmol + 77mmole = 154 mOsmole

1 mmole of dextrose produces 1 mmole of particles (dextrose onle) because dextrose does not dissociate.

Therefore numbere of mOsmoles/L produced by 277 mmoles of dextrose = 277 mOsmoles.

Total mOsmoles = 277mOsmoles/L + 154 mOsmoles/L = 431mOsmoles/L