A lost hiker is found wandering in Joshua Tree National Park

A lost hiker is found wandering in Joshua Tree National Park. He reports that he was lost for two days; he spent most of his time sitting in the shade. He had only two liters of water when he departed his campsite 48 hours previous to being found. He lost about 4 liters of water per day due to evaporative water loss (respiration and sweat) but drank only two liters of water. Calculate the water loss and its effect on average osmolality of body fluids. (Assume your patient has a total body volume of 42L, an ECF volume of 14L and a plasma volume of 3L. Further assume average osmolality of 300 mOsm; the loss of water was distributed among all fluid compartments.)

Calculate the change in fluid osmolality if the hiker had not taken water with him and had lost 4L per day while siting in the shade

I do not understand how to solve this problem. Can someone explain the steps???

Calculate the change in fluid osmolality if the hiker had not taken water and had lost 8L per day because he walked around in the heat of the day instead of resting in the shade.

Thank you.

Solution

Water lost per day = 4 liters per day

total water lost = 4 L x 2 = 8 L

Water intake: 2L

Net water loss = 8L -2L = 6L

Total body fluid before loss (V1) = 42 + 14 +3 = 57L

Total body fluid after loss (V2) = 57-6 L= 51L

Osmolality before the loss was (M1): 300mOsm

The osmolality after loss was M2 :?

M1V1=M2V2

M2= (300 x 57) / 51

=> M2 = 335.3 mOsm

Therefore, the change in the osmolality= 335.3-300 = 35.3 mOsm

b). If the hiker would have not taken any water

M1V1=M2V2

M2= (300 x 57) / 49

=> M2 = 348.9 mOsm

Therefore, the change in the osmolality= 348.9-300 = 48.9 mOsm