As you will hopefully see from the answers here, we may have to make some careful decisions when balancing energy intake, fluid intake, route of administration and concentrations to be used.
QUESTION 1
Show that 10% w/v glucose will be unable to provide the target glucose infusion rate with this limited amount of fluid available.
- 10% glucose is 10 g in 100 ml of solution. To run this at 60 ml/kg/day will give 6 g/kg/day of glucose.
- In a day there are 24 x 60 minutes = 1440 minutes/day
- 6 g/kg/day (6000 mg/kg/day) divided by 1440 minutes = 4.17 mg/kg/minute, whereas at least 6 mg/kg/minute are required.
QUESTION 2
Given that the molecular mass of glucose is 180.18 g/mol, and 1 mole/litre of a non-dissociating solute (such as glucose) will produce a solution with an osmolarity of 1 Osm/L, calculate the osmolarity of 10% glucose.
- 10% glucose is 10 g/100 ml = 100 g/litre
- 100 g divided by the molecular mass of 180.14 g/mole gives an amount of 0.555 moles/litre = 555 mmoles/litre = 555 mOsm/L. This is consistent with being double the concentration of 5% glucose and is also roughly double the osmolarity of plasma. It is also below the range of osmolarity, which is considered to be likely to cause thrombophlebitis despite being slightly hypertonic.
QUESTION 3
Since we only have peripheral access for this child, what is the maximum concentration of glucose we could consider running peripherally if we choose a conservative limit of 600 mOsm/L for the osmolarity?
- 600 mOsm/L = 0.6 Osm/L, which would be the equivalent of 0.6 moles of glucose in a litre of solution
- 0.6 moles of glucose has a mass of 0.6 x 180.14 g/mole = 108.084 g
- 108.084 g/litre = 10.8084 g/100 ml = 10.8%w/v glucose; not much more than the 10% we have available.
QUESTION 4
If we were to fit the minimum desirable glucose delivery rate into the maximum available volume, what would be the concentration and osmolarity of the required solution?
- The minimum rate of glucose delivery is 6 mg/kg/minute, which if running for one whole day (1440 minutes) would require a total amount of glucose of 8640 mg = 8.64 g
- Since the available volume is 60 ml/kg/day we need a concentration of 8.64 g/60 ml = 14.4 g/100 ml = 14.4%w/v
- 14.4g/100 ml = 144 g/litre; 144 g/litre divided by 180.14 g/mole = 0.799 moles/litre = 799 mOsm/L
- This is close to the maximum suggested osmolarity, and higher than many clinicians may be prepared to use peripherally. Additionally, in practice 15% is the closest available solution that is likely to be available without having to dilute down solutions on the ward, and this will have a higher concentration and thus be potentially more irritant.