<Dl> <Dd> C H 2 O = 4 ml / min − 140 mOsm / L 280 mOsm / L × 4 ml / min = 2 ml / min (\ displaystyle C_ (H_ (2) O) = 4 \ (\ mbox (ml / min)) - (\ frac (140 \ (\ mbox (mOsm / L))) (280 \ (\ mbox (mOsm / L)))) \ times 4 \ (\ mbox (ml / min)) = 2 \ (\ mbox (ml / min))) </Dd> </Dl> <Dd> C H 2 O = 4 ml / min − 140 mOsm / L 280 mOsm / L × 4 ml / min = 2 ml / min (\ displaystyle C_ (H_ (2) O) = 4 \ (\ mbox (ml / min)) - (\ frac (140 \ (\ mbox (mOsm / L))) (280 \ (\ mbox (mOsm / L)))) \ times 4 \ (\ mbox (ml / min)) = 2 \ (\ mbox (ml / min))) </Dd> <P> Free water clearance can be used as an indicator of how the body is regulating water . A free water clearance of zero means the kidney is producing urine isosmotic with respect to the plasma . Values greater than zero imply that the kidney is producing dilute urine through the excretion of solute - free water . Values less than zero imply that the kidney is conserving water (likely under the influence of antidiuretic hormone, ADH), resulting in the production of concentrated urine . </P>

How does free-water clearance change in the presence of adh
find me the text answering this question