A. Given a piece of healthy tissue weighing 500 grams, density of the tissue is

Question

A. 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 in mL?

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

Explanation / Answer

According to the given data,

Weight of water (in the healthy tissue) = 60% of 500 gms = 300gms.

Since density of water is 1 g/ml, the volume occupied by water in the tissue = 300 ml.

Volume occupied by intracellular fluid (ICF) = 2/3rd of 300 ml = 200 ml

Volume occupied by extracellular fluid (ECF) = 1/3rd of 300 ml = 100 ml

Given the osmolarity of both ICF and ECF = 300 milliosmols/L.

Also, the tissue is placed in 2000ml of solution containing 8.775 gm of NaCl. Therefore the solution contains approx. 150 millimoles of NaCl. Hence, osmolarity of the solution = 300 milliosmols/L. (since NaCl dissociates into two particles in solution).

A. There will not be any uptake of fluid inside the tissue. Therefore the volume of intracellular fluid will remain same, i.e. 200 ml.

B. The final volume of extracellular fluid in this tissue (including the NaCl solution) = 2100 ml

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.