Answer
A 9 year old female patient (weight = 80.5 lbs) is admitted to the hospital for diabetic ketoacidosis and new onset type 1 DM.
The physician writes:
- 0.9% NaCl 10ml/kg/hr
- Insulin Regular IV 0.3 units/kg/hr
i) What rate of administration (ml/min) would you instruct the nurse to give the 0.9% NaCl
(2 marks)
Weight:
1 lbs = 0.45 kg (You need to remember this conversion ratio)
80.5 lbs = 36.225 kg
0.9% NaCl (Normal Saline) 10ml/kg => 362.25ml per hour
362.25/60 = 6.04 ml/min
ii) After a few hours the physician orders 1000 ml of ½ NS and D5W with 40 mmol KCl to be given at 80ml/hr. At what rate in (mmol/kg/hr) is potassium being administered?
(2 marks)
There is 40 mmol of KCl in 1000ml.
Therefore in 80 ml of the solution there is:
40 X (80/1000) = 3200/1000 = 32/10 = 3.2 mmol KCl => administration rate is 3.2 mmol KCl / hour.
Pt weight is 36.225 kg
3.2/36.225 = 0.0883 mmol/kg
Answer: 0.0883 mmol/kg/hr
iii) What is the osmolality of the 1000 ml ½ NS + D5W with 40 mmol KCl?
MW [g/mol]: KCl = 74.5; NaCl = 58.4; Dextrose = 180
(4 marks)
- Osmosis is the spontaneous diffusion of solvent from a solution of low solute concentration into a more concentrated one through a semipermeable membrane.
- The amount of osmotically active particles in a solution is expressed in osmoles (or milliosmoles).
- Osmotically active particles are either molecules or ions, independent of charge
- For substances that maintain their molecular structure when they dissolve (e.g. glucose), osmolarity and molarity are the same.
- For substances that dissociate when they dissolve, the osmolarity is the number of ‘free’ particles times the molarity. That is a 1 molar solution of NaCl would be 2 osmolar (1 osmolar for Na+ plus 1 osmolar for Cl−).
- Osmolarity: number of osmoles per litre of solution (eg. mOSm/L)
- Osmolality: number of osmoles per kilogram of solvent. (eg. mOsm/kg)
- Osmolarity is calculated.
- Osmolality is not calculated, it is measured in clinical laboratories using an osmometer.
- Values of osmolality and osmolarity (at low concentrations) for a solution are similar. So you can calculate the osmolarity and then estimate that is also what the osmolality will be. State the final answer in mOSm/kg.
In the solution the osmotically active particles present are: KCl, NaCl and Dextrose.
One mole of KCl will disassociate into K+ and Cl-, resulting in 2 osmoles of active particles. The same applies to NaCl. Dextrose does not disassociate, 1 mole of dextrose will still give 1 osmole of active particle.
1000 ml of ½ NS + D5W means 0.45% sodium chloride and 5% dextrose in water.
- Within which also contains 40 mmol of KCl.
NaCl:
0.45% NaCl is 0.45 g in 100ml water. (Normal Saline NS is 0.9%)
0.45g/100ml
4.5g/1000ml
No. of moles in 4.5g of NaCl is: 4.5/58.4 => 0.07705 mol/L = 77.05 mmol/L
No. osmoles NaCl is 77.05 x 2 => 154.11 mOsm/L
KCl
No. osmoles of KCl is 40mmol x 2 = 80 mOSm/L
Dextrose
D5W is 5% dextrose in water, ie. 5 g dextrose in 100ml water.
5g/100ml => 50g/1000ml
No. moles dextrose in 50g is 50/180 = 0.278 mol => 278 mmol
1 mole of dextrose = 1 osmole dextrose
=> 278 mOsm/L
Osmolarity of the solution is: 154 + 80 + 278 = 512 mOsm/L
As it is not possible to calculate osmolality, I will state that osmolality is roughly the same as the osmolarity value.
Answer: Osmolality of the solution is approximately 512 mOsm/kg. (Note the units)
References:
Aulton's Pharmaceutics E-Book: The Design and Manufacture of Medicines (p. 44-45). Elsevier Health Sciences.
Khan Academy Molarity vs Osmolarity