Given a piece of healthy tissue weighing 500 grams, density of the tissue is 1 g

Question

Given a piece of healthy tissue weighing 500 grams, density of the tissue is 1 gram/mL and consisting of 60% water by weight. Also, given that the water in this tissue is distributed such that 2/3 of the water is intracellular and 1/3 is extracellular and that both the ICF and ECF have an osmolarity = 300 milliosmols/L. If you place this tissue in 2 liters of solution containing 8.775 grams of NaCl, what will be the final volume of intracellular fluids? _________ mL

   What will be the final volume of the extracellular fluid in this tissue (including the NaCl solution)? _________ mL

Explanation / Answer

First calculate the volume of tissue -

Density = weight/volume

Volume = weight/ density

Volume of tissue = 500/1 = 500 ml or 0.5 L

As tissue is 60% water therefore total water content of tissue =(0.5×60)/100 = 0.3 L

Osmolarity of 2L solution containing 8.775 g of Nacl

Thus water will flow from extracellular to intracellular, compartment as extracellular will have more Osmolarity. The flow of fluid will stop when both extracellular and intracellular fluid will have same Osmolarity -

Therefore Osmolarity gained = 75 milliosmol/L by both compartment.

Final intracellular fluid volume = 0.2+1 =1.2 L or 1200ml

Final extracellular fluid volume = 1100ml

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(A very simple approach to solution of this problem would be just to add 1 L to the initial ICF and ECF volume as each would eventually gain equal volume of water as they will have same final Osmolarity)

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