Question

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb?

Answer 1

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         $180 \frac{mg}{kg} \times 71.215 kg$

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = $\frac{0.65 g}{100 g}$

                           = $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

                           = $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.