Question

For a caffeinated drink with a caffeine mass percent of 0.51% and a density of 1.00 g/mL, how many mL of the drink would be required to reach an LD50 of 170 mg/kg body mass if the person weighed 171 lb?

Answer 1

  (1) convert body mass in lb to kg 

211 lb = (211 lb)*(1 kg/2.2046 lb) = 95.71 kg 

(2) calculate the amount of caffeine required for a person of that mass at the LD50 

(150 mg/kg)*(95.71 kg) = 14356 mg 

(3) convert % (w/w) to % (w/v) 

0.61% (w/w) = 0.61 g/100 g = (0.61 g)/((100 g)/(1.00 g/mL)) = 0.61 g/100 mL 

(4) calculate the volume that contains the amount of caffeine calculated 

14356 mg = 14.356 g = (14.356 g)/(0.61 g/100 mL) = 2353 mL

Answer 2

Answer:

2,585.47 mL of the drink would be required to reach an LD-50 for 171 lb person.

Explanation:

LD-50 of caffeine = 170 mg/kg body mass

Mass of the person = 171 lb = 77.56 kg

1 lb = 0.453592 kg

Lethal dose for person with mass 171 lb = 77.56 kg × 170 mg/kg =13,185.92 mg

13,185.92 mg = 13.18592 g

(1 mg = 0.001 g)

Percentage of caffeine in drink = 0.51 %

Mass of drink = M

$0.51%=\frac{13.18592 g}{M}\times 100$

M = 2,585.47 g

Volume of the drink = V

Density of the drink = D = 1.00 g/mL

$Density=\frac{Mass}{Volume}$

$1.00=\frac{2,585.47 g}{V}$

$V=\frac{2,585.47 g}{1.00 g/mL}=2,585.47 mL$

2,585.47 mL of the drink would be required to reach an LD-50 for 171 lb person.