test1-review - 2 1 More drug unbound More drug bound Assume...

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Assume a man has a hematocrit of 0.4, is 70% water by weight and weighs 100 kg. He has consumed 35 grams of alcohol (which distributes evenly in total body water). Assume the water content of his plasma is 95% and of his red blood cells (RBCs) is 65%. What is his blood plasma alcohol level , the amount of alcohol in his cells and his blood alcohol level? His total body water = 70L. This is the V d of EtOH. Therefore the concentration of alcohol in his body water is 35 g/ 70L = 500 mg/L (or 0.5 mg/ml). If his hematocrit is 0.4 then 100 ml of his blood is composed of 60 ml of plasma and 40 ml of cells. The 60 ml of plasma is 95% water (therefore 57 ml water). The 40 ml of RBCs is 65% water (therefore 26 ml water). So, 100 ml of blood contains 83 ml of water (57ml + 26ml). If there is 0.5 mg EtOH per ml of water and 57 ml water per 60 ml plasma, there would be 57ml x 0.5mg = 28.5 mg EtOH per 60 ml plasma. This equals 47.5 mg per 100 ml of plasma ( his plasma alcohol level ). In RBCs, there is 26ml x 0.5mg = 13mg EtOH per 40 ml cells. This equals 32.5 mg per 100 ml
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test1-review - 2 1 More drug unbound More drug bound Assume...

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