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Physiology · General physiology: osmolarity calculation
Sample itemA 55-year-old man has serum sodium 140 mmol/L, glucose 180 mg/dL and blood urea nitrogen 28 mg/dL. Using osmolarity = 2 x Na + glucose/18 + BUN/2.8, what is his calculated serum osmolarity in mOsm/L?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
C. 300
The sodium term is 2 x 140 = 280. Glucose contributes 180/18 = 10, and urea contributes 28/2.8 = 10. The total is 280 + 10 + 10 = 300 mOsm/L, which is slightly above the normal 285 to 295 range. Sodium and its accompanying anions account for most plasma osmolarity.
Option by option
A. 290
290 results if the urea term is left out (280 + 10).
B. 280
280 results if only the sodium term is counted and glucose and urea are ignored.
C. 300 (correct)
280 + 10 + 10 = 300 mOsm/L.
D. 310
310 results if BUN is divided by 1.4 instead of 2.8 (giving 20), which is a wrong conversion factor.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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