4. The body of a man contains 49.5 L of water. He drinks 0.5 L of pure water. Su

Question

4. The body of a man contains 49.5 L of water. He drinks 0.5 L of pure water. Suppose that all of this water is absorbed, at first with negligible excretion. What is the percentage decreasen osmotic concentration of the body fluids? Given an initial concentration o water, what is the absolute decrease in concentration? 41. 5 L tho 1.6.= 0.01 010,100 101 1300-mWm1_ . If 3 mmol of sodium moves from cells into each kg of extracellular water, accompanied by 27 g of water, and if the initial extracellular concentration of sodium is 150 mmol/kg water, what is the final concentration of sodium? 5. Daj 50

Explanation / Answer

Question 4-

We will use formula-

C1V1 = C2V2

Here C is concentration and V is volume.

Now, we know that C1 = 300 mosmol/Kg

V1 = 49.5L

V2=49.5 + 0.5L =50L

Therefore by putting all values in the formula-

300 x 49.5 = 50 x C2

C2 = (300 x 49.5) / 50

C2 = 297 mosmol/Kg

Therefore, the absolute decrease in the concentration = 300- 297

The absolute decrease in the concentration = 3 mosmol/Kg

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