Data Analysis Question 1 Consider a 70 kg individual, with a total body water co
ID: 192387 • Letter: D
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Data Analysis Question 1 Consider a 70 kg individual, with a total body water content of 39 liters. The intracellular compartment (ICF) contains 25 liters of water and the extracellular compartment (ECF) has 14 liters. Initially, the osmolarity of both the ECF and ICF is 300mOsm/liter This person now drinks 300 ml of a solution of CaCz (0.333 M) and all 300 ml is completely absorbed from the gastrointestinal tract into the ECF compartment. Assume that the CaCl solution is evenly distributed throughout the ECF compartment and that the cell membranes are completely impermeable to Ca and C Part 1. What would be the gsmolarity of the ECF compartment prior to osmosis occurring? Part 2. Next, assume that osmosis has occurred and osmotic equilibrium has once again been attained. What will be the "new osmolarity of both the ECF and ICF compartments, after water has moved and a new steady state equilibrium is attained?Explanation / Answer
Part - 1) First if all, we will find the new osmolarity of total body water (TBW).
Since, old TBW osmoles = 39L x 300mOsm/L = 11700mOsm
New TBW volume = 39L + 0.3 L = 39.3L
New TBW osmolarity = 11700mOsm/39.3L = 297.7mOsm/L
Next, we will find new ECF volume
Since, old ECF osmoles = 14L x 300mOsm/L = 4200mOsm
Therefore, new ECF volume = 4200mOsmole /297.7mOsm/L = 14.11L
Therfore, osmolarity of ECF prior to osmosis = 4200mOsm/14.11L = 297.66mOsm/L
Part 2) Gain = 0.333moles x 2 = 0.666osmoles = 666mOsmoles
Old ECF osmoles = 14L x 300mOsm/L = 4200mOsm
New ECF osmoles = 4200 mOsm + 666 mOsm = 4866 mOsm
New ECF volume = 4866mOsm/297.7mOsm/L = 16.34L
New osmolarity of ECF = 4866mOsm/ 16.34 = 297.7mOsm/L
Old ICF osmoles = 25L x 300mOsm/L = 7500mOsm
New ICF volume = 7500mOsm/297.7mOsm/L = 25.19 L
New osmolarity of ICF = 7500mOsm/25.19L = 297.7mOsm/L
It is therefore concluded that osmolarity of ECF and ICF is equal in steady state.
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