A patient has the following volumes of distribution: deuterium oxide: 40 liters

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.

1. What is the osmolarity of her body fluids? 285 mOsM

2. How much solute (osmoles) exists in each of the above compartments? 11.4 osmoles in total body water, 3.705 osmoles in ECF, 7.56 osmoles in ICF, 0.798 in plasma.

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.)

4. Patient ingests 0.18 moles of NaCl, all of which is absorbed into the body. Use dissociation constant of 2 for NaCl in all problems.

5. Patient receives an intravenous infusion of 2 L of 0.9% NaCl solution .

6. Patient eliminates 0.5 L of sweat that contains 100 milliosmoles NaCl.

7. Patient ingests 1 L of water containing 0.3 moles of urea, all of which is absorbed.

8. 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.

Here is the Initial box with the initial condition.

Total

ECF

ICF

Plasma

Vol L

40

13

27

2.8

Sol osmol

11.4

3.705

7.56

0.798

mOsM

285

285

285

285

4.

Total

ECF

ICF

Vol L

Sol osmol

mOsM

Total

ECF

ICF

Plasma

Vol L

40

13

27

2.8

Sol osmol

11.4

3.705

7.56

0.798

mOsM

285

285

285

285

Explanation / Answer

1. 1 OsM/ 1.86 degrees = x OsM/0.53 (Freezing point depression being 1.86 degrees)

x (osmolarity)= 0.53 / 1.86= 0.285 OsM= 285 mOsM

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