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a man wt 70 kg , NaCl intake is 200 mmol/kg. excretion of NaCl is as follow
day 1.....30 mmol/kg
day2.......90 mmol/kg
day3 ......180mmol/kg
day4......200mmol/kg
assuming 1lt of 0.9%saline containing 150mmol of NaCl and wt 1 kg. how much will this pt in kg at the end of day 4
a.66 kg
b .68
c. 70
d.72
e.74
plz explain your answer too
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asa bro
yes your answer is right but how to solve this i even don't know but i find this qq on usmle oreintation step 1 qq bock 3 qq no 40/41
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i agree, D
day 1: retains 170mmol/kg
day 2: retains 110mmol/kg
day 3: retains 20 mmol/kg
day 4: retains 0
so, 170 + 110 + 20 + 0 = 300mmol / 150mmol = 2
since 150mmol wt 1kg the pt wt is 72kg
that's the way i did it, correct me if i'm wrong
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First add the fluid the patient lost in 4 days:
30+90+180+200=500
then the intake for 4 days is 200X4=800
The amount retained by the patient is 800-500=300mmol
150mmol------->1kg
300-------------->xkg
x=2kg
Pt wt is 70 and gain 2 more, so 72kg at day4
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bilal1& macusa is it a bolus 200mmol or 200mmol/day???
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i guess 200mmol/day,
is it right, bilal1?