A healthy adult drinks 2 liters of water rapidly. After equilibration, the kidneys excrete the excess water. Given urine flow of 10 mL/min, urine osmolality of 80 mOsm/kg, and plasma osmolality of 290 mOsm/kg, what is the free water clearance?

• A Approximately +7.2 mL/min ✓
• B Negative value indicating water conservation
• C Zero (no free water excretion)
• D Approximately +10 mL/min

Explanation

Free water clearance (CH2O) = V x (1 - Uosm/Posm). Here V = 10 mL/min, Uosm = 80, Posm = 290. CH2O = 10 x (1 - 80/290) = 10 x 0.724 = +7.2 mL/min. B positive value indicates net free water excretion, consistent with water diuresis. The formula shows that when urine is hypotonic relative to plasma, free water is being cleared from the body.

Reference: Guyton and Hall Textbook of Medical Physiology, 14th ed.

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