Chapter U4.79, Problem 3E

(a)

Interpretation Introduction

Interpretation:

The number of cans of a regular soft drink that should be consumed by a 65 kg person in a short period of time before exceeding the lethal dose of caffeine must be calculated.

Concept Introduction :

LD₅₀is known as lethal dose which is the dose in mg/kg required to kill the 50 % of living cells.

Answer to Problem 3E

A 65 kg person can drink 364 cans of regular soft drink in a short period of time before exceeding the lethal dose.

Explanation of Solution

Lethal dose (LD₅₀) of caffeine is 140 mg/kg. So for a 65 kg person the amount required will be:

  $\text{140×65 mg of caffeine=9100 mg of caffeine}$

There are 25 mg of caffeine in a can of regular soft drink.

Number of cans of regular soft drink which will contain 9100 mg of caffeine will be:

  $N=\frac{9100}{25}\text{ cans}=\text{364 cans}$

(b)

Interpretation Introduction

Interpretation:

Toxicity of caffeine in moles/kg must be calculated.

Concept Introduction :

Toxicity of a substance in moles/kg means the amount of the substance required in mole per kg body weight of the living being to kill 50 % of living cells.

Answer to Problem 3E

The toxicity of caffeine is $7.2\times {10}^{-4}\text{moles/kg}$ .

Explanation of Solution

Molecular weight of caffeine is 194.

LD50 of caffeine is 140 mg/kg.

  $140mg=0.140g$

Moles of caffeine in 140 mg can be calculated as follows:

  $\begin{array}{l}n=\frac{m}{M}\\ =\frac{0.140\text{ g}}{194\text{ g/mol}}\\ =7.2\times {10}^{-4}\text{ mol}\end{array}$