Question

deuterium oxide: 42 liters            

inulin: 13 liters                           

Evans blue: 2.8 liters

Her plasma shows a freezing point depression of 0.55 degrees C. For an ideal solution, a 1 OsM solution depresses the freezing point 1.86 degrees C.

1. osmolarity of her body fluids? --> 0.295

2. How much solute (osmoles) exists in each of the above compartments? (shown below)

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. You should be able to give the numerical values for volume and osmolarity of the ECF and ICF. Briefly explain your rationale for solving each problem. (Treat each solution below as a separate problem; do not make them additive.)

3. Patient receives an intravenous infusion of 1.5 L of glucose solution containing 0.22 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.

4. Patient ingests 1 L of water containing 0.4 moles of urea, all of which is absorbed.

Total Plasma Voll 42 2.8 ECF ICF 1329 3.85 8.58 296 296 Sol osmol 12.43 0.83 296 mOsM 296 Total Voll Sol osmol mOsM Total VOLL Sol osmol mOsM

Answer 1

Deuterium oxide = Total body water = 42 L

Inulin = ECF volume = 13 L

ICF = 29 L

1 Osm = 1000 mOsm

1000 mOsm depresses freezing point by 1.86 °C.

0.55° C depression in freezing point = 0.55/1.86 * 1000 = 295.7 mOsm

So Osmolarity of plasma = Osmolarity of ECF= Osmolarity of ICF = 295.7 mOsm/L

2)Solutes in TBW = 295.7 * 42 = 12.4 Osm

Solutes in ICF = 295.7* 29 = 8.58 Osm

Solutes in ECF = 12.4 - 8.58 = 3.82 Osm

3)Total osmoles of glucose = 0.22

Volume of TBW = 43.5 L

Number of osmoles in ICF= 1/3*0.22 = 0.073

Volume of ECF = 13+ 1.5= 14.5 L

Osmolarity of ECF = 3.82 + 0.147/14.5 = 3.967/14.5 = 273.8 mOsm/L

Osmolarity of ICF = 8.58 + 0.073/29= 8.653/29= 298.3 mOsm/L