Answer :
Final answer:
The total osmolarity of the solution is 134.5 mOsmol/L. The 20 ml vial of ammonium chloride injection contains 93 mEq of ammonium chloride.
Explanation:
To calculate the osmolarity of a solution, we need to find the number of moles of each compound, then multiply by the number of particles each compound will dissociate into, and convert to milliosmoles. The 2% Dextrose (C6H12O6) is 2g/100ml. Therefore, in 750 mL we have 15g of dextrose. As the molecular weight of Dextrose is 180, the moles in the solution are 15/180 = 0.0833 moles. Since Dextrose does not dissociate, the osmolarity contribution from dextrose is 0.0833*1000 = 83.3 mOsmol/L.
The 0.2% Sodium Chloride (NaCl) is 0.2g/100ml. Therefore, in 750 mL, we have 1.5g of salt. Given the molecular weight of NaCl is 58.5, the moles of NaCl are 1.5/58.5 = 0.0256 moles. Since NaCl dissociates into 2 particles (Na+ and Cl-), the osmolarity contribution from the salt is 0.0256*1000*2 = 51.2 mOsmol/L. Therefore, the total osmolarity of the solution is 83.3+51.2 = 134.5 mOsmol/L.
Regarding the Ammonium Chloride injection, 20 ml of injection contains 250mg/ml * 20 = 5000 mg = 5g of NH4Cl. The number of moles is 5/53.5 = 0.093 moles. The milliequivalents (mEq) are found by multiplying the moles of solute by the valence (1 for NH4+), and scaling by 1000, giving 0.093*1*1000 = 93 mEq.
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