Question

A patient has the following volumes of distribution:

deuterium oxide: 40 liters

inulin: 13 liters

Evans blue: 2.8 liters

Her plasma shows a freezing point depression of 0.53 degrees C. For an ideal solution, a 1 OsM solution

depresses the freezing point 1.86 degrees C.

The osmolarity calculated is 0.285 OsM.

Using the data above as the initial conditions in each case, determine the effects of the following procedures on

the water and solute in the ECF and ICF after equilibrium has occurred. Briefly, explain your rationale for solving each

problem.

1. Patient receives an intravenous infusion of 1.5 L of glucose solution containing 0.25 moles of glucose per liter

of solution. Calculate ECF and ICF volumes and osmolarity at the point where 1/3 the glucose has entered the

cells.

	Total body	ECF	ICF
Solute
Volume
Osmolarity

Answer 1

1.All exchange of water and solutes with in the external environment occur through the ECF (eg:intravenous infusionand intake or loss via the gastriintestinal tract).Changes in the ICF are secondary to fluids shifts between the ECF and ICF.

• Except a brief periods of seconds to minutes, the ICF and ECF are in osmotic equilibrium

The volumes of various body fluid compartments can be estimated in the normal adult as:

• Total body water(TBW)=0.6xbody weight
• EXtracellular fluid volume(ECF)=0.2xbody weight
• Intracellular fliuid volume(ICF)=0.4xbody weight
• Plasma volume(P)=0.25xECF volume
• Interstitial fluid volume(IF)=0.75xECF volume

Suppose here the patient is weighing abou 50kg and receives intravenous infusion of of 1.5 litre of glucose solution as 5%of dextrose that means the patient is receiving 0.25moles of glucose per litre of solution.This solution is an essentially iso-osmotic plasma.

Here we will see the ECF,ICF and osmolarity(osmol) at the point of 1/3 the glucose has entered the cells

Equations given previously

This is an isotonic solution,and will therefore remain confined initially to the ECF.The ECF volume will increase by 500ml,and ICF volume will be unchanged

• Total body water comprises of 60% of body weight.Of this,40% is in the ICF and 20% is in the ECF.Then total body water(0.6x50kg)=30L
• ICF volume =0.4xbw,ie,0.4x50kg=20L
• ECF volume =0.2xbw+500ml,ie,0.2x50kg+500ml of solution=10+500ml

Therefore ECF=10.5L

Molarity,denoted by unit 'M' the number of moles of a solute dissolved in one liter of solution.It can be written as mol/l.The formula(equation) for osmolarity is:

• Osmolarity(osmol)=Number of moles of dissolved particles/Number of litres of solution

At the point of 1/3 the glucose has entered in to the body,

Osmolarity(osmol)=0.125 moles of glucose/500ml of glucose solution,ie,0.25

	Total body	ECF	ICF
Solute	5%Dextrose	10.5L	20L
Volume	500ml
Osmolarity	0.25m