mahendra
Forum Guru
Topics: 173
Posts: 419
|
A 23-year-old man has an intracellular fluid volume of 28 L,
an extracellular fluid volume of 14 L, a plasma volume of 3 L,
and an extracellular fluid osmolarity of 285 mOsm/L. The man drinks
3 L of water and consumes 10 mEq of sodium (in the form of potato chips)
. What is his approximate extracellular osmolarity (assuming osmotic
equilibrium)?
A. 266 mOsm/L
B. 285 mOsm/L
C. 291 mOsm/L
D. 300 mOsm/L
E. Cannot be determined
|
| asmi
Forum Hero
Topics: 1043
Posts: 4,609
|
i guess it should remain same B
MAY BE IAM WRONG, Mahender plz do explain
|
| Idiopathic
Forum Guru
Topics: 19
Posts: 641
|
It should go down, so it should be A. 1 mEq of Na+ corresponds with 20 mL of pure H20, right? So 10 mEq = 200mL/3000mL = 0.67%...this would be like giving hypotonic saline. Sound good? I am kind of winging the last part, but I go with A.
|
| mahendra
Forum Guru
Topics: 173
Posts: 419
|
u got it 
|
| asmi
Forum Hero
Topics: 1043
Posts: 4,609
|
thanks for the explanation 
I need to work hard man :oops: ,i didn't knew it.
|
| asmi
Forum Hero
Topics: 1043
Posts: 4,609
|
but you said 200/3000=67%.......can you complete the calculation further
|
| Idiopathic
Forum Guru
Topics: 19
Posts: 641
|
my math is off a decimal point or so...200/3000 = ~6.7%. This is the way to think about it:
1 mEq of Na+ requires 20 mL of H20 to be isotonic. Any fluid above that 20mL/mEq lowers the plasma osmolality. In this case, we add 2800 mL extra H20 that does not have salt associated with it. So, our plasma osmolality has to go down.
I got the 1 mEq Na+/20 mL water from renal, because that is how much 'free' water is created at the TAL of the loop when that pump takes back Na+/K+/2 Cl-...
|
| asmi
Forum Hero
Topics: 1043
Posts: 4,609
|
thanks
|
|
| |
| | | | | | | | |
