Question

Be sure to answer all parts. Fluoride ion is poisonous in relatively low amounts: 0.2 g of F− per 70 kg of body weight can cause death. Nevertheless, in order to prevent tooth decay, F− ions are added to drinking water at a concentration of 1 mg of F− ion per L of water. How many liters of fluoridated drinking water would a 70−kg person have to consume in one day to reach this toxic level?

Answer 1

Answer:

200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.

Explanation:

Toxic level of fluoride ions = 0.2 g per 70 kg body weight

Concentration of fluoride ions in drinking water = 1 mg/ L = 0.001 g/L

1 mg = 0.001 g

Let the volume of drinking water consumed by person to reach the toxic level of fluoride concentration be V.

For 70 kg body weight 0.2 grams of fluoride ions will be toxic.

$V\times 0.001 g/L=0.2 g$

$V=\frac{0.2 g}{0.001 g/L}=200 L$

200 liters of fluoridated drinking water would have to be consumed by as person to reach the toxic level.