Question

The driver of a vehicle involved in a collision was detained by police under suspicion of DIU and blood was drawn half hour later. He declared he equally split a bottle of wine with his partner about an hour before the collision.

a) If he is 190 pounds, what would be the BAC expected? (3p)

b) If his actual BAC was 0.15 g/dL, calculate the BAC at the time of collision. (3p)

c) How much wine should he had to drink to reach that value as Cmax? (3p)

useful info:

Volume distribution: Vd = 0.7 L/kg men and Vd = 0.6 L/kg women

Rate of elimination: k = 0.015 g/100 ml/h men and k = 0.016 g/100 ml/h women

Retrograde analysis at (Widmark): Ct = BAC + k(time sampling –t) with t = time driver stopped

Density: 20C = 0.789 g/ ml of ethanol

Standard drink: 14g, beer regular 5vol %, whiskey 86 proof, vodka 80 proof, wine 14 vol%.

Plasma/blood: 1.15 1 fl.oz = 29.6 ml

1 kg = 2.2 pounds

Answer 1

1 BAC = (Alcohol consumed in grame) x 100 Body weight ingroms x ) x100 MI DUAL CAMERA REDMI NOTE 5 PRO as in wine, concentration is 14 voll. so, in half bottle; i.o, in 325 ml, (45.5 ml of ethanal is present, 1. 45.5 ml x 6.789 g/ml (Density - mall = 135.89.gm . His body weight in grams = KD 86182gms. For men, r=0.68. X loo .. BAC = 35.89 8618280.68 = 10.061 This is BAC as percentage. i.l, before accounting for time. So, after shes BAC = (BAC as "%) - (Time in hes + 0.015) = (0.061) - (1 + 0.015) = 0.053

(a) so after half and houk= 0.061 Vb) During coltisona 0.053. (b) During collison, if BAC is o.159/dL 1.0, 0.001591ml. so, at the time of collison, 0.0015 + ( 13 x 0.015) -0.009